4.81 ml is taken from a stock solution of concentration 0.800 m and added to a 10.0 ml volumetric flask and diluted up to the calibration mark.
What is concentration of a solution?
The amount of solute present in a specific volume of a solution is its concentration. It can be formulated in a variety of ways. The "percentage approach" is the most used way to express a solution's concentration.
Molarity (M) = moles of solute ÷ liters of solution
Here, the concentration of the solution is 0.800 M and the volume of solution is 4.81 ml ( = 0.00481 liter )
So,
0.800 = moles of solute ÷ 0.00481
Or, moles of solute = 0.003848 mole
Now, 4.81 ml solution is poured in a 10 ml ( = 0.01 leter) volumetric flask and diluted up to the calibration mark.
Here, the volume of the solution = volume of the flask
And the moles of solute remains same.
So,
Concentration (M)= 0.003848 ÷ 0.01
= 0.3848
The new concentration will be 0.3848 M.
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Potassium metal reacts with chlorine gas to form solid potassium chloride. Answer the following:
Write a balanced chemical equation (include states of matter)
Classify the type of reaction as combination, decomposition, single replacement, double replacement, or combustion
If you initially started with 78 g of potassium and 71 grams of chlorine then determine the mass of potassium chloride produced.
The balanced chemical equation between pottasium and chlorine is as follows: 2K + Cl₂ → 2KCl. It is a combination reaction.
What is a chemical reaction?A chemical reaction is a process, typically involving the breaking or making of interatomic bonds, in which one or more substances are changed into others.
According to this question, a chemical reaction occurs between potassium metal and chlorine gas to form pottasium chloride as follows:
2K + Cl₂ → 2KCl
The chemical reaction is a combination reaction because it involves the combination of two elements to form a compound.
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Determine the percent yield if 0.0419 mol of ethyl acetate is produced but 0.0722 mol is calculated as the theoretical yield?
First DEFINE and THEN Compare and contrast the three states of matter. Draw a picture on the
molecular level for each.
There are mainly three types of state of matter and they solid, liquid as well as gas. Gas molecule has no any attraction force on the other hand solid molecule has strong attraction force.
There are four different states of matter: solids, liquids, gases, as well as plasma. Adding or subtracting heat energy from a substance can frequently modify the state of matter in that substance. For instance, adding heat may cause ice to melt into liquid water as well as water to scald.
A condition of a substance known as gas entirely conforms to the form of the container it is placed in despite having no shape of its own. Solids have a defined shape as well as volume, whereas liquids just have a defined volume as well as no shape, and gases lack both shape and volume.
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A 50. 6 grams sample of magnesium hydroxide (Mg(OH)2) is reacted with 45. 0 grams of hydrochloric acid (HCl). What mass of MgCl2 is produced?
82.67 grams of MgCl₂ are produced when 50.6 grams of Mg(OH)₂ and 45.0 grams of HCl are reacted.
The balanced chemical equation for the reaction between magnesium hydroxide and hydrochloric acid is:
Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
To find the mass of MgCl₂ produced, we need to determine which reactant is limiting. This can be done by calculating the number of moles of each reactant and comparing them to the stoichiometric ratio in the balanced equation.
Number of moles of Mg(OH)₂ = 50.6 g / 58.32 g/mol = 0.868 mol
Number of moles of HCl = 45.0 g / 36.46 g/mol = 1.235 mol
According to the balanced equation, 1 mole of Mg(OH)₂ reacts with 2 moles of HCl. Therefore, Mg(OH)₂ is the limiting reactant, since only 0.868 moles of Mg(OH)₂ are available to react with HCl.
From the balanced equation, we know that 1 mole of Mg(OH)₂ produces 1 mole of MgCl₂. Therefore, the number of moles of MgCl₂ produced is also 0.868 moles.
The molar mass of MgCl₂ is 95.21 g/mol. Therefore, the mass of MgCl₂ produced is:
Mass of MgCl₂ = 0.868 mol x 95.21 g/mol = 82.67 g
Therefore, approximately 82.67 grams of MgCl₂ are produced when 50.6 grams of Mg(OH)₂ and 45.0 grams of HCl are reacted.
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Help please!
An Arrhenius acid increases the concentration of hydrogen ions when dissolved in solution.
True
False
Answer:
The answer is true
Explanation:
An Arrhenius acid is a substance that increases the concentration of hydrogen ions in solution when it is dissolved in water.
Answer:
false
Explanation:
A 2.00 cm3 sample of metal is found to weigh 4.54 g. Calculate the density of the metal in units of g/cm3 and g/mL. Be sure to answer all parts
Answer:
2.27 g/cm³
Explanation:
The formula for density is D = m/v
Using this, we can calculate the density by inputting
m = 4.54 g
V = 2.00 cm³
D = 4.54/2.00
= 2.27
= 2.27 g/cm³
Therefore, the density of the metal is 2.27 g/cm³
Hope this helps!! :)
Answer:
Density = Mass/Volume
= 4.54g * 2cm^3
= 2.27g/cm3
= 2.27g/ml
mL and Cm^3 are same
Place the particles in order from smallest to largest.
1Electron
2Molecule
3Neutron
4Nucleus
5 atom
Answer:
smallest to largest:
Electron, Neutron, Atom, Molecule, Nucleus
Explanation:
sorry if it's not right
Electron--Neutron--Nucleus-- atom--Molecule is the order from smallest to largest.
Why electron is smaller?Electron is the smallest subatomic particles having the radius is about 9.1× 10−31 kg while on the other hand, the size of neutrin is 1.7×10−15 meters. Both neutron and proton are present in the nucleus so nucleus is bigger than neutron. Nucleus is present inside an atom so atom is bigger than nucleus whereas atom is smaller than molecule.
So we can conclude that Electron--Neutron--Nucleus-- atom--Molecule is the order from smallest to largest.
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What is the mass of 9.3 x 1024
molecules of glucose, C6H12O6?
(C6H12O6; 180.18 g/mol)
[?] g C6H₁2O6
Mass (g)
Enter
Answer:Molecular mass of 9.3 x 10^ 24 molecules of glucose = 1675.6 x 10^ 24
It is given that the mass of glucose is 180.18 g/mol
so if we have to calculate the mass of 9.3x 10^ 24 molecules of glucose
we will apply a simple unitary method i.e,
9.3 x 10^ 24 * 180.18 g/mol = 1675.67 x 10^ 24
To calculate the molecular mass of a molecule, multiply the subscript (number of atoms) by the atomic mass of each element in the molecule and add those masses together.
Remember that - To determine the compound's molecular mass in grams per mole, use the molecular formula.
Divide the supplied mass by the molar mass of the chemical to convert it to moles.
By dividing the number of moles by Avogadro's number, you may convert from moles to molecules.
Explanation:
The mole concept is used here to determine the mass of glucose. The mass of glucose with 9.3 x 10²⁴ molecules is 2.78 × 10⁴⁹ g.
What is a mole?One mole of a substance is defined as that amount of it which contains as many particles or entities as there are atoms in exactly 12 g of Carbon - 12.
The equation which is used to calculate the number of moles is given as:
Number of moles (n) = Given mass / Molar mass
Number of molecules = Number of moles of molecules × 6.022 × 10²³
Number of moles = 9.3 x 10²⁴ / 6.022 × 10²³
= 1.54 × 10⁴⁷
Mass = Number of moles × Molar mass
= 1.54 × 10⁴⁷ × 180.156
= 2.78 × 10⁴⁹ g
Thus the mass of glucose is 2.78 × 10⁴⁹ g.
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2
Which represents a mixture?
*
Ос.
Explanation:
A mixture is an impure substances with the following properties:
Composition is indefinite i.e they consist of two or more elements and or compounds in any proportion by massConstituents retain their identities i.e physical properties are retainedConstituents reacts differently to changed conditions. They can easily be separated into constituents by physical methodsExamples are air, sea - water, muddy water, palm -wine, petroleum, alloys etc.
Using this description it should be easy to identify a mixture.
For the following unbalanced equation: Cu(s) + AgNO₃(aq) ➞ Cu(NO₃)₂(aq) + Ag(s) if 1.50 moles of copper and 3.00 moles of silver nitrate are available to react, how many grams of silver are produced? *
4 points
430 g Ag
324 g Ag
302 g Ag
108 g Ag
A suitable recrystallizing solvent is one in which the chemical to be purified is relatively ____ at low temperature and completely ____ at or near the boiling point of the solvent.
A suitable recrystallizing solvent is one in which the chemical to be purified is relatively insoluble at low temperature and completely soluble at or near the boiling point of the solvent.
What are the solubility characteristics required for a recrystallizing solvent?Recrystallization is a widely used technique for purifying solid compounds. It involves dissolving the impure compound in a suitable solvent at an elevated temperature and then allowing it to slowly cool down, causing the compound to recrystallize in a purer form. The choice of solvent is crucial in this process, as it directly affects the success of the purification.
A suitable recrystallizing solvent should exhibit two key characteristics. Firstly, it should be relatively insoluble or have low solubility in the compound to be purified at low temperatures. This ensures that the impurities remain dissolved while the target compound precipitates out during the cooling process. Secondly, the solvent should be completely soluble in the compound at or near its boiling point. This ensures that the compound fully dissolves during the initial dissolution step, allowing the impurities to be left behind.
By carefully selecting a recrystallizing solvent with these solubility characteristics, it is possible to obtain highly pure crystals of the desired compound. The choice of solvent depends on various factors such as the nature of the compound, the desired purity level, and the availability of suitable solvents.
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describe what happens to carbon dioxide in the carbon cycle
In the carbon cycle, carbon dioxide (CO₂) undergoes various processes, including photosynthesis, respiration, and combustion, resulting in its release and uptake by different components of the Earth system.
Carbon dioxide plays a crucial role in the carbon cycle, which is the natural process through which carbon is exchanged and recycled in the biosphere, geosphere, hydrosphere, and atmosphere. Here are the main processes involving carbon dioxide in the carbon cycle:
1. Photosynthesis: Plants, algae, and some bacteria use sunlight, water, and carbon dioxide during photosynthesis to produce glucose (C₆H₁₂O₆) and oxygen (O₂).
2. Respiration: Organisms, including plants, animals, and microorganisms, release carbon dioxide as a byproduct of cellular respiration. During respiration, glucose and oxygen are consumed to produce carbon dioxide, water, and energy. This process returns carbon dioxide to the atmosphere.
3. Combustion: When organic matter, such as fossil fuels (coal, oil, and natural gas), wood, or biomass, is burned, carbon dioxide is released into the atmosphere.
4. Absorption and Dissolution: Carbon dioxide can dissolve in bodies of water, including oceans, lakes, and rivers. In the oceans, it undergoes a series of chemical reactions to form carbonic acid (H₂CO₃), bicarbonate ions (HCO₃⁻), and carbonate ions (CO₃²⁻). This process, known as ocean acidification, affects marine ecosystems.
5. Carbon Sequestration: Some carbon dioxide is absorbed and stored by plants and inorganic carbon sinks such as soil, sediment, and geological formations.
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the complete question is:
What is the role of carbon dioxide in the carbon cycle and how does it participate in the process?
1. A 0.5100-g sample of CaCO3 is dissolved in 12 M HCl and the resulting solution is diluted to 250.0 mL in a volumetric flask.
a. How many moles of CaCO3 are used (MW=100.1 g/mol)? .5100g x (1 mol/100.1g)= 0.005095 mols
b. What is the molarity of the Ca^2+ in the 250 mL of solution? .005095 mols/.250L= 0.02038M
c. How many moles of Ca^2+ are in a 25.0-mL aliquot of the solution in b? .02038M x 0.025L= 5.095 x 10^-4 2.
a)the number of moles of CaCO3 used is approximately 0.005096 mol. b) the molarity of solution is approximately 0.02038 M. c) the number of moles of the solution is approximately 5.095 × 10^(-4) mol.
a. To find the number of moles of CaCO3 used, we need to divide the given mass by the molar mass of CaCO3.
Given:
Mass of CaCO3 = 0.5100 g
Molar mass of CaCO3 = 100.1 g/mol
Number of moles of CaCO3 = Mass / Molar mass
Number of moles of CaCO3 = 0.5100 g / 100.1 g/mol
Number of moles of CaCO3 ≈ 0.005096 mol (rounded to 4 decimal places)
b. To determine the molarity of Ca^2+ in the 250 mL of solution, we need to use the number of moles of CaCO3 and the volume of the solution.
Given:
Number of moles of CaCO3 = 0.005096 mol
Volume of solution = 250.0 mL = 0.250 L
Molarity of Ca^2+ = Number of moles / Volume
Molarity of Ca^2+ = 0.005096 mol / 0.250 L
Molarity of Ca^2+ ≈ 0.02038 M
c. To calculate the number of moles of Ca^2+ in a 25.0-mL aliquot of the solution, we can use the molarity of Ca^2+ obtained in part (b) and the volume of the aliquot.
Given:
Molarity of Ca^2+ = 0.02038 M
Volume of aliquot = 25.0 mL = 0.025 L
Number of moles of Ca^2+ = Molarity × Volume
Number of moles of Ca^2+ = 0.02038 M × 0.025 L
Number of moles of Ca^2+ = 5.095 × 10^(-4) mol
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here there we go now yall can answer
McCray, Shy'Anne
00
5 of 18
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Finish
Nasha left a hot bowl of soup in a cool room. Her friend tells her that her soup will get cold if she forgets to eat it. How cold could it get?
Answer:
The temperature of soup will decrease.
The soup will get as cold as the room temperature.
The temperature of soup will decrease. The soup will get as cold as the room temperature.
What is room temperature?Room temperature is a range of air temperatures that most people prefer for indoor settings.
Room temperature is defined differently in different places. It generally refers to a range somewhere between 68 and 74 degrees Fahrenheit.
The temperature of soup will decrease. The soup will get as cold as the room temperature.
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a terminal alkyne was treated with nanh2 followed by propyl iodide. the resulting internal alkyne was treated with ozone followed by water, giving only one type of carboxylic acid. provide a systematic, iupac name for the internal alkyne.
The systematic Iupac name for the internal alkyne is : 4-octyne
When only one type of carboxylic acid is made the internal alkyne will be symmetric and the compounds can be 3-hexane or 4-octyne but given that one of the reactants is propyl iodide. Therefore
The Internal alkyne will have an Iupac name 4-octyne
Hence we can conclude that the systematic Iupac name for the internal alkyne is 4-octyne
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What is the atomic number of an atom that has 5 neutrons and 4 electrons
The atomic number is 4.
The atomic number of the given atom is equal to 4.
What is the atomic number?The atomic number (symbol Z) of a chemical element can be defined as the charge number of an atomic nucleus. For an ordinary atom, the atomic number is equal to the number of protons found in the nucleus of an atom of that element.
The atomic number helps to uniquely identify ordinary chemical elements. In an ordinary uncharged atom or neutral atom, the atomic number of that atom is also equal to the number of electrons.
For a neutral atom, the sum of the atomic number and the neutron number gives the atomic mass. The protons and neutrons have approximately the same mass. Atoms that have the same atomic number but different numbers of neutrons or different mass numbers are known as isotopes.
Given the number of electrons in the atom is 4. Therefore, the atomic number of that atom will be equal to 4.
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Determine the number of moles of sodium in 3.20 moles of sodium hydrogen phosphate.
The formula of sodium hydrogen phosphate is Na₂HPO₄.
3.2 moles of Na₂HPO₄ contains 6.40 moles of Sodium atoms.
Two mole of sodium, one moles of hydrogen, one mole of phosphate, and four moles of oxygen make up one mole of sodium hydrogen phosphate, or Na₂HPO₄.
Given;
1 mole of Na₂HPO₄ contains 2 moles of Sodium
3.2 moles of Na₂HPO₄ will contain =?
= 3.2 moles x 2
= 6.40 moles
Therefore;
3.2 moles of Na₂HPO₄ contains 6.40 moles of Sodium atoms.
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Given that sodium bicarbonate dissociates to form Na+
and HCO3- when mixed with water, which of
these would be part of the explanation for taking bicarbonate
(NaHCO3) for excess stomach acid?
When bic
Bicarbonate (HCO3-) would be part of the explanation for taking bicarbonate (NaHCO3) for excess stomach acid.
When mixed with water, sodium bicarbonate dissociates into Na+ and HCO3-. Bicarbonate acts as a base that reacts with the acid in the stomach to neutralize it, causing carbon dioxide gas and water to be produced as by-products. This makes sodium bicarbonate an effective antacid for treating heartburn and other forms of acid reflux.The neutralization reaction can be written as follows:NaHCO3 + HCl → NaCl + CO2 + H2O
Where NaHCO3 is sodium bicarbonate, HCl is hydrochloric acid, NaCl is sodium chloride, CO2 is carbon dioxide, and H2O is water. Bicarbonate neutralizes the acid in the stomach, reducing symptoms of heartburn and other types of acid reflux.
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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.
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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what could happen if an alkaline developer is used in dye penetrant inspections
If an alkaline developer is used in dye penetrant inspections, it can cause the dye to wash out, making it difficult or impossible to detect any flaws or defects in the surface being inspected.
The alkaline developer can also react with the dye, altering its chemical properties and making it ineffective for future inspections.
This can lead to inaccurate or incomplete inspections, which can have serious consequences if the surface being inspected is critical for safety or performance.
It is important to always use the correct type of developer for the specific dye penetrant being used to ensure accurate and reliable results.
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Match up the characteristics below with the type of molecular bond they describe. Bonds found in Halite (between Na+ and Cl-) Bonds found between Si and O in the Si-O tetrahedron Bonds inside the water molecule (between the H and O ) Bonds that exist between two water molecules Strongest bond type Weakest bond type Bonds that are used by water to dissolve sal
The characteristics and the type of molecular bond they describe:
1. Bonds found in Halite (between Na⁺ and Cl⁻): Ionic bond
2. Bonds found between Si and O in the Si-O tetrahedron: Covalent bond
3. Bonds inside the water molecule (between the H and O): Covalent bond
4. Bonds that exist between two water molecules: Hydrogen bond
5. Strongest bond type: Covalent bond
6. Weakest bond type: Van der Waals bond
7. Bonds that are used by water to dissolve salt: Ionic bond
The ionic bond is a type of molecular bond found in halite (between Na⁺ and Cl⁻). The Si-O tetrahedron is held together by a covalent bond. The bond inside the water molecule (between the H and O) is also a covalent bond. The hydrogen bond is the type of bond that exists between two water molecules. The covalent bond is the strongest bond type, while the van der Waals bond is the weakest bond type. Water uses the ionic bond to dissolve the salt.
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if a 1.0 l flask is filled with 0.22 mol of n2 and 0.22 mol of o2 at 2000°c, what is [no] after the reaction establishes equilibrium? (kc
The equilibrium concentration of [NO] is 0.246 mol/L.
To find the concentration of NO ([NO]) at equilibrium, we need to use the equilibrium constant (Kc) and the initial concentrations of N2 and O2.
Given that, Initial volume (V) = 1.0 L
Initial moles of N2 = 0.22 mol
Initial moles of O2 = 0.22 mol
Equilibrium constant (Kc) = 0.10
Using the equation: N2 + O2 ⇌ 2NO
Let's denote the change in concentration of NO as 'x'.
Since the stoichiometry of the balanced equation shows that 1 mol of N2 reacts to form 2 mol of NO, the change in concentration of NO is 2x.
The initial concentration of N2 and O2 remains the same since their moles and volume are constant.
At equilibrium, we can express the concentrations as follows:
[N2] = (0.22 mol) / (1.0 L) = 0.22 mol/L
[O2] = (0.22 mol) / (1.0 L) = 0.22 mol/L
[NO] = (2x mol) / (1.0 L) = 2x mol/L
According to the equilibrium constant expression, Kc = [NO]^2 / ([N2] * [O2])
0.10 = (2x)^2 / (0.22 * 0.22)
Solving this equation will provide the value of x, which corresponds to the equilibrium concentration of NO.
After solving the equation, we find that x ≈ 0.123 mol/L.
Therefore, [NO] at equilibrium is approximately 2x = 2 * 0.123 = 0.246 mol/L.
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In a catalyzed reaction a reactant is often called a:.
Answer:
it is called a substrate
Explanation:
hopfully this helps u
Answer:
The reactant in an enzyme-catalyzed reaction is called a substrate
PLEASE HELP AND GET 100 POINTS!
Burning fuels can pollute the air.
Give TWO substances which pollute the environment when fuels burn.
Answer:
When fossil fuels are burned, they release nitrogen oxides into the atmosphere, which contribute to the formation of smog and acid rain. ... When fossil fuels are burned, they release nitrogen oxides into the atmosphere, which contribute to the formation of smog and acid rain.
how did the ancient greek understanding of atomic particles likley influence dalton research?
Answer:
The Greeks called these particles atomos, meaning indivisible, and the modern word “atom” is derived from this term. Democritus proposed that different types and combinations of these particles were responsible for the various forms of matter.
Explanation:
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The idea of indivisibility of the atom upheld by Dalton was largely derived from early Greek philosophers.
John Dalton was the first scientist to begin an empirical study of atomic theory. Before this time, all that was known about the atom were largely based on the speculation of Greek philosophers.
However, one recurring theme among Greek philosophers is that atoms are indivisible. This idea that atoms are indivisible became a very strong pivot of the Dalton's theory of the atom.
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Which of these statements is one of the conclusions that formed the basis of dalton’s atomic theory?.
The correct answer is option D.
The statement that formed the basis of Dalton’s atomic theory is atoms are the smallest particles of matter and cannot be divided farther.
In 1808, Dalton presented a theory named atomic theory which suggests that atoms are the smallest particles of an element and it is impossible to divide them further.
According to his theory every element is composed of these tiny particles.
Furthermore, his theory suggests that atoms neither can be divided nor destroyed.
In a particular matter, for example gold, all atoms have similar properties while their mass varies for every single different element.
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The complete question is:
Which of these statements is one of the conclusions that formed the basis of Dalton’s atomic theory?
a. Atoms can only change into atoms of another element through nuclear reactions.
b. Atoms of gases have less mass than atoms of liquids and solids.
c. Atoms of a particular element all have the same number of protons.
d. Atoms are the smallest particles of matter and cannot be divided farther.
please help thank you
Based on the Hund's Rule, the three electrons in a 2p orbital are arranged as shown in option C.
What is an orbital?The term orbital refers to a region in space where there is a high probability of finding the electron. We know that an electron in an atom could be defined by the use of four sets of quantum numbers which are;
The principal quantum numberThe orbital quantum numberThe spin quantum numberThe magnetic quantum numberHund's rule states that, electrons occur singly before pairing takes place. In this case, we can see that the only arrangement that obeys the Hund's Rule is option C.
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For a given reaction with ΔH = -28.1 kJ/mol, the ΔG = 0 at 366 K. The value of ΔS must be _______ J/K-mol, assuming that ΔH and ΔS do not vary with temperature.
To determine the value of ΔS, we can use the equation relating ΔG, ΔH, and ΔS is ΔG = ΔH - TΔS. Given that ΔG is zero at 366 K, we can rewrite the equation as 0 = ΔH - (366 K)ΔS
Given that ΔG is zero at 366 K, we can rewrite the equation as 0 = ΔH - (366 K)ΔS,
Rearranging the equation to solve for ΔS:
ΔS = ΔH / (366 K),
Substituting the given value of ΔH as -28.1 kJ/mol:
ΔS = (-28.1 kJ/mol) / (366 K)
Since the units of ΔH are in kJ/mol and we need the answer in J/K-mol, we convert kJ to J:
ΔS = (-28.1 × 10^3 J/mol) / (366 K)
Calculating the value: ΔS ≈ -76.5 J/K-mol
S is favoured whereas H is unfavourable. If H is negative and S is positive, G will always be negative, making the reaction spontaneous at all temperatures.
Therefore, assuming that ΔH and ΔS do not vary with temperature, the value of ΔS for the given reaction is approximately -76.5 J/K-mol.
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The oxygen in water behaves as though it’s , and the hydrogens behave as though they’re
The oxygen in water behaves as though it’s electronegative, and the hydrogens behave as though they’re electropositive. This is due to the difference in electronegativity between oxygen and hydrogen.
Oxygen is more electronegative than hydrogen, which means that it has a stronger attraction for electrons. As a result, the electrons in a water molecule spend more time around the oxygen atom than they do around the hydrogen atoms.
The oxygen in water is said to be electronegative, while the hydrogens behave as if they're electropositive. This is due to the disparity in electronegativity between the two atoms. Oxygen is more electronegative than hydrogen, implying that it has a stronger attraction for electrons.
Electronegativity is a measure of an atom's ability to attract electrons. It helps to explain why the electrons in a water molecule spend more time around the oxygen atom than they do around the hydrogen atoms.
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