. n specifies fill in the blank 4 l specifies fill in the blank 5 ml specifies fill in the blank 6 ... a.the orbital orientation. b.the energy and average distance from the nucleus. c.the subshell - orbital shape.
Answers
n specifies B. The energy and average distance from the nucleus, l specifies C. The subshell - orbital shape, ml specifies A. The orbital orientation.
n is known Principal quantum number
l is known Azimuthal quantum number
ml is known Magnetic quantum number
The energy and usual distance from the nucleus are described by the fundamental quantum number, or n.
The subshell and orbit's shape are described by the orbital quantum number, or l.
The orbit's direction is described by the magnetic quantum number, ml.
So, n specifies B. The energy and average distance from the nucleus.
l specifies C. The subshell - orbital shape.
ml specifies A. The orbital orientation.
Thus, n specifies B., l specifies C., ml specifies A.
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Related Questions
Which example is NOT an organism?
Answers
Answer:
Option D
Lung
Explanation:
Lung is an Organ not an Organism
-TheUnknownScientist 72
Answer:
lung
Explanation:
the lung is not a life form but an organ. but human, bacteria and mold are all living organisms
Which of these statements relating to ecological succession is true?
During succession, there is no change to the physical or chemical environment.
During succession, existing species resist interaction with new species.
During succession, new species move into an area and colonize it.
Most ecological successions occur over 10 to 15 years.
Answers
Answer: During succession, new species move into an area and colonize it.
Explanation: Ecological succession refers to the process of change in the composition and structure of an ecosystem over time. It occurs due to the interactions between the biotic (living) and abiotic (non-living) components of an environment. As succession progresses, new species gradually establish and thrive in the area, leading to a change in the species composition. This process can occur over a long period of time, ranging from decades to centuries, depending on various factors such as environmental conditions and the specific type of succession.
Help needed plsssssss
Answers
The mass of the water vapor is 13.73 g.
What is the law of conservation of mass?The law of conservation of mass is a fundamental principle in physics and chemistry that states that the total mass of a closed system remains constant during a chemical reaction or physical change. This means that the mass of the reactants in a chemical reaction must equal the mass of the products that are formed.
In other words, the mass of the products that are formed in a chemical reaction cannot be created or destroyed, but only converted from one form to another.
Knowing that;
Total mass of the reactants = Total mass of the products
5 + 10 = 1.27 + x
x is the amount of the water vapor
x = 15 - 1.27
= 13.73 g
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A large fish tank is initially filled with 30 litres of fresh water. You begin to fill the tank by slowly pouring in water with salt concentration of 35 grams per litre (approximate salinity of sea water) at a rate of 2 litres per minute. At the same time, the (perfectly mixed) fluid in the tank is drained from the bottom at a rate of 1 litre per minute. 1. Determine the volume of water in the tank at time t. [1 mark] 2. Let S(t) denote the amount of salt in the fish tank at time t in grams. Show that S(t) satisfies the ODE S
′
(t)=70−
t+30
S
. Write down the appropriate initial condition for the ODE as well. [2 marks] 3. What order is this ODE? Is it linear? Is it separable? [1 mark] 4. Solve the initial value problem to find S(t) using the method of integrating factors. [3 marks] 5. What is the salt concentration in the tank as t→[infinity] ? [1 mark] Part B: Double tanks Next you hook up two fish tanks in a loop so that there is a pipe from tank A to tank B, and also a pipe from tank B back to tank A. Two pumps are added so that you can control the flow rate in each pipe. Initially tank A contains 80 litre of fresh water and tank B 60 litres of fresh water. You begin to pour salt water with concentration 35 grams per litre into tank A at a rate of 2 litres per minute. To keep the tanks from overflowing, you set your pumps so that water is flowing at a constant rate of 4 litres per minute from tank A to tank B, and 2 litre per minute from tank B to tank A. You also put a drain in tank B so that fluid is draining at a rate of 2 litres per minute. 1. Sketch a diagram of the tank setup with arrows for flows entering and leaving each tank. [
1 mark]
2. Let P(t) and Q(t) denote the amount of salt in tank A and tank B respectively. Show that P and Q satisfy a system of ODE's in the form of
P
′
(t)
Q
′
(t)
=c
1
P(t)+c
2
Q(t)+c
3
=c
4
P(t)+c
5
Q(t)
where c
1
,c
2
,c
3
,c
4
and c
5
are constants. Determine the constant c
1
,c
2
,c
3
,c
4
,c
5
and write down appropriate initial conditions. [2 marks] 3. Show that the system of ODE's can be converted into the following second order ODE for P(t) P
′′
(t)=−
60
7
P
′
(t)−
600
1
P(t)+
3
14
State the initial conditions for this ODE. [2 marks] 4. Solve this second order ODE to find P(t), and hence Q(t) as well.
Answers
1. The volume of water in the tank at time t is given by the equation
Volume(t) = 30 + t.
2.The appropriate initial condition for the ODE is S(0) = 0, as there is no salt initially in the tank.
3. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.
4. The solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^{2/2} + 30t)} dt) / e^{(t^{2/2} + 30t)})\)
5. the salt concentration in the tank as t→infinity is zero.
1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained. The tank is being filled at a rate of 2 liters per minute and drained at a rate of 1 liter per minute.
Since the tank starts with an initial volume of 30 liters, the volume of water in the tank at time t can be calculated using the equation:
Volume(t) = Initial volume + (Rate of filling - Rate of draining) * t
Volume(t) = 30 + (2 - 1) * t
So, the volume of water in the tank at time t is given by the equation
Volume(t) = 30 + t.
2. Let S(t) denote the amount of salt in the fish tank at time t in grams.
To show that S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t),
we need to take the derivative of S(t) with respect to t and substitute it into the given ODE.
Taking the derivative of S(t), we have:
S'(t) = 0 - (1+0)S(t) + 0
S'(t) = -S(t)
Substituting this into the given ODE, we get:
-S(t) = 70 - (t+30)S(t)
Simplifying the equation, we have:
S'(t) = 70 - (t+30)S(t)
Therefore, S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t).
The appropriate initial condition for the ODE is S(0) = 0,
as there is no salt initially in the tank.
3. This ODE is a first-order linear ordinary differential equation. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.
4. To solve the initial value problem for S(t) using the method of integrating factors, we first rewrite the ODE in standard form:
S'(t) + (t+30)S(t) = 70
The integrating factor is given by:
\(\mu(t) = e^{(\int (t+30) dt)} = e^{(t^2/2 + 30t)\)
Multiplying both sides of the equation by μ(t), we have:
\(e^{(t^2/2 + 30t)} * S'(t) + e^{(t^2/2 + 30t)} * (t+30)S(t) = 70 * e^{(t^2/2 + 30t)\)
Applying the product rule to the left side of the equation, we get:
\((e^{(t^{2/2} + 30t) * S(t))' = 70 * e^{(t^{2/2} + 30t)})\)
Integrating both sides of the equation with respect to t, we have:
\(\int (e^{(t^2/2 + 30t)} * S(t))' dt = \int (70 * e^{(t^2/2 + 30t))} dt\)
Using the fundamental theorem of calculus, the left side becomes:
\(e^{(t^2/2 + 30t)} * S(t) = \int (70 * e^{(t^2/2 + 30t))} dt\)
Simplifying the right side by integrating, we get:
\(e^{(t^2/2 + 30t)} * S(t) = 70 * \int e^{(t^2/2 + 30t)} dt\)
At this point, the integration of \(e^{(t^2/2 + 30t)\) becomes difficult to express in terms of elementary functions.
Hence, the solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^2/2 + 30t)} dt) / e^{(t^2/2 + 30t)\)
5. As t approaches infinity, the exponential term \(e^{(t^2/2 + 30t)\) becomes very large, causing the salt concentration S(t) to approach zero. Therefore, the salt concentration in the tank as t→infinity is zero.
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The salt concentration in the tank as t approaches infinity is 70/3.
1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which water is being drained from the bottom.
At a rate of 2 litres per minute, water is being poured into the tank. So after t minutes, the amount of water poured into the tank is 2t litres.
At a rate of 1 litre per minute, water is being drained from the tank. So after t minutes, the amount of water drained from the tank is t litres.
Since the tank was initially filled with 30 litres of fresh water, the volume of water in the tank at time t is given by:
Volume(t) = 30 + 2t - t
Volume(t) = 30 + t
2. Let S(t) denote the amount of salt in the fish tank at time t. To determine the ODE for S(t), we need to consider the salt being poured into the tank and the salt being drained from the tank.
The salt concentration in the water being poured into the tank is 35 grams per litre. So the amount of salt being poured into the tank per minute is 35 * 2 = 70 grams.
The amount of salt being drained from the tank per minute is S(t)/Volume(t) * 1.
Therefore, the ODE for S(t) is:
S'(t) = 70 - S(t)/Volume(t)
The initial condition for this ODE is S(0) = 0, since there was no salt in the tank initially.
3. The ODE S'(t) = 70 - S(t)/Volume(t) is a first-order linear ODE. It is not separable since the variables S(t) and Volume(t) are mixed together.
4. To solve the initial value problem for S(t), we can rewrite the ODE as:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
This is a linear ODE of the form y'(t) + p(t)y(t) = g(t), where p(t) = 1/Volume(t) and g(t) = 70 * Volume(t).
To solve this type of ODE, we can multiply both sides by an integrating factor, which is the exponential of the integral of p(t).
The integrating factor is exp(integral of 1/Volume(t) dt) = exp(ln(Volume(t))) = Volume(t).
Multiplying both sides of the ODE by the integrating factor, we get:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
Volume(t) * S'(t) + Volume(t) * S(t) = 70 * Volume(t)^2
( Volume(t) * S(t) )' = 70 * Volume(t)^2
Integrating both sides with respect to t, we get:
Volume(t) * S(t) = 70/3 * Volume(t)^3 + C
S(t) = 70/3 * Volume(t)^2 + C/Volume(t)
Using the initial condition S(0) = 0, we can solve for C:
0 = 70/3 * 30^2 + C/30
C = -70000
Therefore, the solution for S(t) is:
S(t) = 70/3 * Volume(t)^2 - 70000/Volume(t)
5. As t approaches infinity, the volume of water in the tank becomes very large. In this case, we can approximate the volume of the tank as t, since the rate at which water is being poured in is 2 litres per minute. So the salt concentration in the tank as t approaches infinity is given by:
S(t)/Volume(t) = (70/3 * t^2 - 70000/t) / t
As t approaches infinity, the second term (-70000/t) approaches 0, so the salt concentration in the tank as t approaches infinity is:
S(t)/Volume(t) = 70/3 * t^2 / t = 70/3 * t
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according to the uncertainty principle, as you localize a wave in time, you become more uncertain about its:_____.
Answers
Answer:
position
Explanation:
cant know both at the same time
4,1,-1,+1/2 element in quantum numbers?
Answers
The element that has the outer shell as has been shown here is gallium.
What element has these quantum numbers?We know that the quantum numbers are the tools that can be used to be able to obtain the position of the electron in the atom. We know that it gives the description of electrons that are present in the atoms of elements.
In this case, we have the quantum numbers that describe the element as 4,1,-1,+1/2 . This shows that the electron must be in a 4p orbital. This must be an element that is in group 13 of the periodic table.
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Aditya Birla Cement Manutacturing Company manufactures cement for use in construction of stone builelings. Beginning work in process inclustes 400 urvits that are 20% compiete with respect to conversion and 30% complete with respect to materials. Ensing work in process inclades 200 units that are 40% complete with respect to corversion and 50 E complete with respect to materials, 2,000 units were stated duting the perlod. Also, assume that $9,900 of material costs and $14,880 of cortversion costs were in the beginning inventory and $180,080 of materials and $409,200 of conversion costs were added to paoduction duing the period. What is the total cost pet equivalent unit using the weighted average method? Multiple Choice $26860 $26785 578000 $26500
Answers
The correct option is $26785.To calculate the total cost per equivalent unit using the weighted average method, we need to consider the costs incurred in both the beginning work in process and the units added during the period.
First, let's calculate the equivalent units of production for both conversion and materials:
Conversion costs:
Beginning work in process: 400 units × 20% complete = 80 equivalent units
Units added during the period: 2,000 units × 40% complete = 800 equivalent units
Total equivalent units for conversion = 80 + 800 = 880 equivalent units
Material costs:
Beginning work in process: 400 units × 30% complete = 120 equivalent units
Units added during the period: 2,000 units × 50% complete = 1,000 equivalent units
Total equivalent units for materials = 120 + 1,000 = 1,120 equivalent units
Next, let's calculate the total costs incurred:
Conversion costs:
Beginning work in process cost: $14,880
Costs added during the period: $409,200
Total conversion costs = $14,880 + $409,200 = $424,080
Material costs:
Beginning work in process cost: $9,900
Costs added during the period: $180,080
Total material costs = $9,900 + $180,080 = $189,980
Now, we can calculate the total cost per equivalent unit:
Total cost per equivalent unit = (Total conversion costs + Total material costs) / (Total equivalent units for conversion + Total equivalent units for materials)
Total cost per equivalent unit = ($424,080 + $189,980) / (880 + 1,120)
Total cost per equivalent unit ≈ $267.85
Therefore, the correct option is $26785.
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What is the mass of solute in a 500mL solutiom of 0. 200 M Sodium Phosphate
Answers
The mass of solute in a 500mL solution of 0.200 M Sodium Phosphate is approximately 16.394 grams.
To find the mass of solute in a 500mL solution of 0.200 M Sodium Phosphate, we can follow these steps:
1. Identify the molar concentration (M) of the solution, which is given as 0.200 M.
2. Convert the volume of the solution from mL to L: 500mL = 0.500L.
3. Calculate the moles of solute (Sodium Phosphate) using the formula: moles = Molarity × Volume. So, moles = 0.200 M × 0.500 L = 0.100 moles.
4. Find the molar mass of Sodium Phosphate (Na3PO4). The molar mass of Na is 22.99 g/mol, P is 30.97 g/mol, and O is 16.00 g/mol. Therefore, the molar mass of Na3PO4 is (3 × 22.99) + 30.97 + (4 × 16.00) = 163.94 g/mol.
5. Finally, calculate the mass of solute using the formula: mass = moles × molar mass. So, mass = 0.100 moles × 163.94 g/mol = 16.394 g.
In summary, the mass of solute in a 500mL solution of 0.200 M Sodium Phosphate is approximately 16.394 grams.
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URGENT!!!!
Part D: Challenge
An unknown mixture is set in beaker on the teacher's table. Describe the five different parts of the
mixture and how we can separate it.
Answers
An unknown mixture is set in beaker on the teacher's table the five different parts of the mixture and we can separate it by method evaporation, distillation, sedimentation, and filtration
Mixture is the composed of one or more pure substances in varying compositions called as mixture and unknown mixture is separated by shake the separatory funnel to allow intimate mixing of the solutions and effect extraction of the compound from the organic mixture and mixture can be physically separated by sedimentation process, filtration, crystallization, distillation, diffusion, panning, magnetic separation, and adsorption are a few of the methods used to separate materials and the substance is said to be pure when a sample of it has purified all other type of material
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PLEASE HELP FIRST GETS BRAINLIEST
Answers
Answer: I'm going to guess and say it is h2 or o2 I am not sure.
Explanation:
Which of the following pictures correctly show the structure of the H+ ion? Click on the picture to see each choice.
Answers
Answer: The answer is the one with the letter P in the red circle.
Explanation:
Answer:
this
Explanation:
because
Define uniform and non-uniform motion.
Answers
Answer:
When a body covers equal distances in a straight line, equal intervals of time are called uniform motion
When a body covers unequal distance in equal intervals of time in a straight line is called non-uniform
Answer:
Uniform motion- is motion in which the
velocity does not change.
Non-uniform motion-occurs when the velocity of an object is not
constant : the object speeds up or slows down during its motion, or changes direction.
Hope it will help you !!
3.
a) Using the five-step method, predict the predominant Bronsted-Lowry acid base
reaction when solutions of perchloric acid and sodium dihydrogen phosphate are
combined.
b) Identify one conjugate acid-base pair from the reaction equation.
c) Predict whether the equilibrium will favour the formation of the reactants or
products. Support and explain your answer.
Answers
Answer:
a) The predominant Brownstead-Lowry acid- base is perchloric acid
b) A conjugate acid base pair is HClO4 and ClO4^-
c) forward reaction is favored and equilibrium position lies towards the far right
Explanation:
The molecular equation is;
HClO4(aq) + NaH2PO4(aq) -----> H3PO4(aq) + NaClO4(aq)
The net ionic equation of the reaction is;
H^+(aq) + H2PO4^-(aq) ----> H3PO4(aq)
The predominant Brownstead-Lowry acid- base is perchloric acid .
A conjugate acid base pair is HClO4 and ClO4^-
Since the both species; perchloric acid and sodium dihydrogen phosphate are both highly soluble in water yielding ions and also dissociate to a large extent in solution, the forward reaction is favoured leading to the formation of phosphoric acid. Remember that phosphoric acid does not easily dissociate in solution hence the forward reaction is faster than the reverse reaction and the equilibrium position lies towards the far right.
what is the role of detergent, ethanol, and salt in the extraction process?
Answers
Detergent, ethanol, and salt can all play important roles in extraction processes, depending on the specific type of biomolecule being extracted and the purification needs.
Detergent, ethanol, and salt can all play important roles in different types of extraction processes. Here's an overview of how each of them can be used:
Detergent: Detergents are often used to extract proteins or other biomolecules from biological samples. Detergents are able to solubilize membrane-bound proteins by disrupting the lipid bilayer of the cell membrane. This allows the proteins to be extracted into a solution. Detergents can also help to break apart protein complexes, making it easier to isolate individual proteins.
Ethanol: Ethanol is commonly used as a solvent in extractions because it can dissolve a wide range of organic compounds, including lipids and proteins. Ethanol can also be used to precipitate DNA or RNA from a solution. When added to a solution of DNA or RNA, ethanol causes the nucleic acids to clump together and form a visible precipitate, which can then be collected and further purified.
Salt: Salt is often used in DNA and RNA extractions to help remove proteins and other impurities from the sample. When salt is added to a solution, it can cause proteins to denature and precipitate out of the solution. The salt can then be removed along with the precipitated proteins. In addition, salt can help to bind DNA or RNA to a solid support, such as a column or a membrane, making it easier to isolate the nucleic acid from other components of the sample.
In summary, detergent, ethanol, and salt can all play important roles in extraction processes, depending on the specific type of biomolecule being extracted and the purification needs. Detergents are useful for solubilizing proteins, ethanol can dissolve a wide range of organic compounds, and salt can help to remove impurities and bind DNA or RNA to a solid support.
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Question 2: if an object is heavy, the acceleration:
Increases
Equal
Decreases
Giving brainlist. On this one
Answers
Answer:
Explanation:
if an object is heavy,the acceleration decreases because the acceleration produced on a body is inversely proportional to its mass.
HELP!!!!!!!!!!!!!Rank the following elements by increasing atomic radius: carbon, aluminum, oxygen, potassium.
Answers
Answer:
oxygen<carbon<aluminium<potassium
Explanation:
atomic radius increases down the group but decreases across the period
A 5. 0 l flask containing o2 at 2. 00 atm is connected to a 3. 0 l flask containing h2 at 4. 00 atm and the gases are allowed to mix. What is the mole fraction of h2? report your answer to two decimal places.
Answers
The mole fraction of hydrogen is 0.55.
The volume of oxygen = 5 litres
The pressure of oxygen = 2 atm
The volume of hydrogen = 3 litres
The pressure of hydrogen = 4 atm
The universal gas constant is R.
R = 8.31 J mol—¹ K—¹
The ideal gas equation is,
PV = nR T
The number of moles of oxygen is,
PV = nRT
\(n = \frac{ PV }{RT }\)
\(n = \frac{2 \times 5}{8.31 \times 273} \)
n = 0.004 moles
The number of moles of oxygen is 0.004 moles.
The number of moles of hydrogen is,
PV = nRT
\(n = \frac{ PV }{RT }\)
\(n = \frac{4 \times 3}{8.31 \times 273} \)
n = 0.005 moles
The number of moles of hydrogen is 0.005 moles.
The mole fraction of hydrogen is,
\(Mole \: fraction = \frac{ Moles \: of \: solute }{Total \: number \: of \: moles \: of \: solute \: and \: solvent}\)
\(Mole \: fraction = \frac{0.005}{0.009}\)
= 0.55
Therefore, the mole fraction of hydrogen is 0.55.
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Consider the following reaction in a gas phase:C(s) + H2O(g) ⇄ CO(g) + H2(g) KC = 0. 2 at 1000 °CCalculate the concentration of CO in an equilibrium mixture (in mol/L) if a reaction mixture initially contains only reactants, and the equilibrium concentration of H2O(g) is [H2O] = 0. 500 M at 1000 °C
Answers
The concentration of CO (g) in the equilibrium mixture is 0.020 M. In other words, only a small amount of CO (g) is produced in this reaction at 1000°C. T 0.020 M.
The concentration of CO in an equilibrium mixture (in mol/L) is 5.8 M.
Given that the concentration of H2O (g) is [H2O] = 0.500 M at 1000°C, and the reaction is:
C(s) + H2O(g) ⇄ CO(g) + H2(g) KC = 0.2 at 1000°C
We need to determine the concentration of CO in an equilibrium mixture (in mol/L)
if a reaction mixture initially contains only reactants.
We can solve this problem using the ICE table method as follows:
Let x be the change in concentration of H2O (g) and CO (g) when they reach equilibrium.
Then the equilibrium concentrations of CO (g) and H2 (g) are equal to x. Hence, the equilibrium concentration of H2O (g) is (0.500 - x) M. Substitute these values in the expression for Kc and solve for x.
Kc = [CO (g)] [H2 (g)] / [H2O (g)] [C (s)]
= 0.2[CO (g)] = Kc [H2O (g)] [C (s)] / [H2 (g)]
= 0.2 × (0.500 - x) / x
We can simplify this expression by cross-multiplication to get:
5x = 0.1 - 0.2xx = 0.02 M
Substituting x = 0.02 M in the expression for [CO (g)], we get:
[CO (g)] = 0.2 × (0.500 - 0.02) / 0.02 = 5.8 M (approx.)
Therefore, the concentration of CO (g) in an equilibrium mixture (in mol/L) is 5.8 M. The problem requires us to find the equilibrium concentration of CO (g) in a mixture that initially contains only reactants.
To solve this problem, we need to use the expression for the equilibrium constant Kc, which is given by:
Kc = [CO (g)] [H2 (g)] / [H2O (g)] [C (s)]
We can also use the ICE table method to solve this problem. In this method, we start with the initial concentration of the reactants and calculate the change in concentration of each species as they reach equilibrium.
We then use the equilibrium concentrations to calculate the value of Kc and solve for the unknowns. Here is how we can set up the ICE table for this problem: Reaction:
C(s) + H2O(g) ⇄ CO(g) + H2(g)
Initial: [C] = [H2]
= 0 M,
[H2O] = 0.500 M
Equilibrium: [C] = [H2] = x,
[H2O] = 0.500 - x,
[CO] = [H2] = x
Change: +x +x -x -x
Substituting the equilibrium concentrations into the expression for Kc, we get:
Kc = [CO] [H2] / [H2O] [C]
= x² / (0.500 - x)
= 0.2
Solving for x, we get: x = 0.020 M
Substituting this value of x into the expression for [CO], we get:
[CO] = x = 0.020 M
Therefore, the concentration of CO (g) in the equilibrium mixture is 0.020 M.
In other words, only a small amount of CO (g) is produced in this reaction at 1000°C. T 0.020 M.
The concentration of CO in an equilibrium mixture (in mol/L) is 5.8 M.
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Materials:
Food labels
Magnifying glass or eye glasses (if necessary)
Procedure:
1. Look for ten (10) food labels that you can find at home (ex. cereal drink, chocolate candy, soy sauce, etc.)
2. You may also find product labels other than food such as medicine, household
cleaning products, cosmetics, and toiletries.
3. List down in the table below, the compounds in the product label and the constituent
elements. There are cases that you will need to look up the constituent elements
because they may not be obvious from the compound name (e.g., citric acid, oil).
Write its uses.Use(s)/Function
4. Careful in handling corrosive substances. Use gloves or wash your hands after holding it.
Compounds and their constituent elements written in the food labels
Food Product
Compound
Constituent Element
Use(s)/Function
Ex. Cereal drink
Iron pyrophosphate
Iron, phosphorus, oxygen
Zinc sulfate
Zinc, sulfur, oxygen
What distinguishes compounds from elements? Applying, Synthesizing
pls help me...:((
Answers
Answer:
Nice experiment
Explanation:
Balance the chemical reaction
using an atom inventory.
What is the coefficient for
sodium?
[?]Na + Cl₂ → [ ]NaCl
Answers
The coefficient for sodium in the balanced equation would be 1.
Balancing chemical equationWhen balancing a chemical equation, you want to make sure that the number of atoms of each element is the same on both sides of the equation. To do this, you can follow these steps:
The unbalanced equation with the correct chemical formulas for the reactants and products.Na + Cl2 -> NaCl
Count the number of atoms of each element on both sides of the equation.On the left side, there is 1 Na and 2 Cl.
On the right side, there is 1 Na and 1 Cl.
Add coefficients to the reactants and/or products to balance the number of atoms of each element. In this case, we can balance the number of chlorine atoms by adding a coefficient of 2 in front of NaCl, like this:Na + Cl2 -> 2 NaCl
Thus, the coefficient of sodium is 1.
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In the Diels Alder reaction between 2,3-dimethyl-1,3-butadiene and maleic anhydride to form cyclic anhydride, what was the %recovery using the information below?
Weight of maleic anhydride used: 1 85 mg = 0.1 85g Volume of diene used: 0.215mL Weight of cyclic anhydride crystals recovered: 0.251g % Recovery:
Answers
Recovery = (Weight of product recovered ÷ Theoretical yield) × 100% Recovery = (0.251 g ÷ 0.27594 g) × 100% Recovery = 91.08%Therefore, the % recovery of the reaction between 2,3-dimethyl-1,3-butadiene and maleic anhydride to form cyclic anhydride was 91.08%.
Diels-Alder Reaction Diels-Alder reaction is a chemical process that joins a conjugated diene with a dienophile (a compound containing a double bond) to form a six-membered ring called a cyclohexene ring. It is a chemical reaction that can be used to make new carbon–carbon bonds. The reaction was discovered by two German scientists, Otto Diels and Kurt Alder, in 1928.In the Diels-Alder reaction between 2,3-dimethyl-1,3-butadiene and maleic anhydride to form cyclic anhydride, the % recovery was calculated using the following data:Weight of maleic anhydride used: 185 mg = 0.185 gVolume of diene used: 0.215 mLWeight of cyclic anhydride crystals recovered: 0.251 g% Recovery:We can calculate the percent recovery using the following formula:% Recovery = (Weight of product recovered ÷ Theoretical yield) × 100The theoretical yield of the product can be calculated from the balanced chemical equation as follows:2,3-dimethyl-1,3-butadiene + Maleic Anhydride → Cyclohexene-1,2-dicarboxylic AnhydrideThe molar mass of 2,3-dimethyl-1,3-butadiene is 68 g/mol. The molar mass of Maleic Anhydride is 98 g/mol. The molar mass of Cyclohexene-1,2-dicarboxylic Anhydride is 146 g/mol.Using these values, we can calculate the number of moles of each reactant: moles of diene used = (0.215 mL)(0.788 g/mL)/(68 g/mol) = 0.00248 mol moles of maleic anhydride used = 0.185 g/98 g/mol = 0.00189 mol The theoretical yield of cyclohexene-1,2-dicarboxylic anhydride = 0.00189 mol × 1 mol/1 mol = 0.00189 molThe theoretical yield of cyclohexene-1,2-dicarboxylic anhydride = 0.00189 mol × 146 g/mol = 0.27594 gNow, we can substitute these values into the percent recovery equation:% Recovery = (Weight of product recovered ÷ Theoretical yield) × 100% Recovery = (0.251 g ÷ 0.27594 g) × 100% Recovery = 91.08%Therefore, the % recovery of the reaction between 2,3-dimethyl-1,3-butadiene and maleic anhydride to form cyclic anhydride was 91.08%.
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Under standard conditions of 298 K and 1 atm, which statement is true?a. diamond converts to graphite spontaneouslyb. graphite converts to diamond spontaneouslyc. none of the above
Answers
Under a standard conditions diamond converts to graphite spontaneously.
The constant of thermodynamic equilibrium K is the correct quotient of species activities in a reaction equilibrium. An activity cannot be many orders of magnitude greater than 1 at ordinary temperature as well as pressure.
Thermodynamic properties are termed as system characteristics that can specify the system's state. Some constants, such as R, do not explain the state of a system and so they are not attributes. Transformation of diamond to graphite is spontaneous as well as advantageous, according to thermodynamics. However, because of kinetics rather than thermodynamics governs this reaction, it proceeds at an exceedingly slow rate. Diamond will be kinetically stable but thermodynamically unstable.
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What is the pOH if the [OH-]= 0.165 M? What is the pH of this basic solution? *Please round your answer(s) to the appropriate number of significant figures. Your answer can be in standard notation or i letter "e" in place of x10.* 1 N
Answers
Answer:
78
Explanation:
Answer:
The OH would be .7825 and the basic solution is a strong base.
Explanation:
What you would do is -log(0.165) in your calculator which would give you 0.7825160065 as an answer. Im not sure what the significant figure is so I will not be rounding to that, but that is your answer for the first part.
The second part: because your pOH is a .7825, this would be consiered a strong base in the pOH because it is closer to 1 which is your base.
tin hydrogenooxolate formula
Answers
Answer:
Tin(IV) Hydrogen Oxalate. Alias: Stannic Hydrogen Oxalate. Formula: Sn(HC2O4)4. Molar Mass: 474.8178. :: Chemistry Applications:: Chemical Elements, Periodic Table.
Explanation:
Answer:
Sn(HC2O4)4 --> formula for Tin(IV) Hydrogen Oxalate
All of the following are anthropogenic greenhouse gases except Group of answer choices a. carbon dioxide. b. molecular oxygen. c. chlorofluorocarbons (CFCs). d. methane.
Answers
Answer:
c
Explanation:
its not a plant based gas
1. Determine the pH of the following solutions:
a. 1 x 10-3 M HCI
Answers
please refer to the attachment
Hope This Helps You ❤️
Answer:
HCI is a strong acid : HCI -> h^+ + CI^- [ H^+ ] = 1.0 x 10 ^-3 -> pH = -log[ H^+ ] = -log 10 ^-3 = 3
Explanation:
Does taking away an electron change the position on the periodic table?
Answers
Answer:
no it doesn't change the position in periodic table
As the atomic number increases along a row of the periodic table, additional electrons are added to the same, outer most shell.
Someone help me please- Im stuck, will give brainliest
Which of the following ions is written correctly?
Select one:
a. Chlorine ion
b. Potassium ion
c. Oxygen ion
d. Sulfur ion
Answers
Answer:
Option b.) Potassium ion, is written correctly.
Explanation:
Hope this helps! :)
Potassium ion is the ion which is written correctly as all other are non-metals which in form of ions must be written with suffix-ide.
What are ions?An ion is defined as an atom or a molecule which has a net electrical charge. There are 2 types of ions :1) cation 2) anion . The cation is the positively charged ion and anion is the negatively charged ion . As they are oppositely charged they attract each resulting in the formation of ionic bond.
Ions consisting of single atom are mono-atomic ions while which consists of two or more ions are called as poly-atomic ions . They are created by chemical interactions . They are very reactive in their gaseous state and rapidly react with oppositely charged ions resulting in neutral molecules.Ions are formed on dissociation.
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what pertinent chemical information is needed in order to determine
if a chemical hazard truly exist?
Answers
In order to determine if a chemical hazard truly exists, there are several pertinent pieces of chemical information that must be considered. Firstly, the chemical composition of the substance in question must be examined, including its molecular formula and structural properties.
This information can help to determine the potential toxicity and reactivity of the chemical, as well as its potential routes of exposure (e.g. inhalation, ingestion, skin contact).
Additionally, data on the physical properties of the chemical - such as its melting point, boiling point, and solubility - can be important in determining how the substance may behave in different environments and under different conditions.
Finally, information on the potential environmental impact of the chemical, such as its persistence in the environment or its potential to bioaccumulate, can also be crucial in assessing the overall hazard posed by the substance.
By considering all of these factors together, a comprehensive picture of the potential hazards associated with a particular chemical can be developed, helping to inform appropriate risk management strategies and protective measures.
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Use your knowledge of atomic calculations to complete the chart.
Element
Protons
Neutrons
Electrons
Atomic
Number
3
15
Atomic
Mass
7
31
35
19
28
31
P
ci
Ni
K
Ag
H
39
19
-18
47
OK
1
1
14
14
06
3
Si
w
74
1010
Ne
10
10
Answers
Answer:
For hydrogen. H
Explanation:
Atomic number is 1
Proton number is 1
Mass number is 2
Neutron number is 1