The correct answer to the question is:
Mg(s) + Zn(NO₃)₂aq) -> Mg(NO₃)₂(aq) + Zn(s) (Option B)
What is a chemical equation?Chemical equations are representations of chemical reactions using symbols and formula of the reactants and products.
The reactants are located on the left side while the products are located on the right side.
Reactants —> Products
The balancing of chemical equations follows the law of conservation of matter which states that matter can neither be created nor destroyed during a chemical reaction but can be transferred from one form to another.
What is complete ionic equation?This is an equation showing all the substance present in the reaction in ionic form
What is net ionic equation?This is an enhanced equation of the complete ionic equation where the spectator ions (i.e ions present on reactant and product side) are not included.
Considering the net ionic equation given from the question:
Mg(s) + Zn²⁺ + 2NO₃⁻ → Mg²⁺ + 2NO₃⁻ + Zn(s)?
Thus, we can conclude that the correct answer to the question is:
Mg(s) + Zn(NO₃)₂aq) -> Mg(NO₃)₂(aq) + Zn(s) (Option B)
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Complete question
What has a complete ionic equation of
Mg(s) + Zn2+ + 2NO3 → Mg2+ + 2NO3 + Zn(s)?
O A. Mg(s) + Zn2+ → Mg2+ + Zn(s)
O B. Mg(s) + Zn(NO3)2(aq) → Mg(NO3)2(aq) + Zn(s)
O C. Mg(NO3)2 + Zn(s) → Zn2+ + 2NO3 + Mg(s)
O D. Mg(s) + 2Zn(NO3)(aq) → 2MgNO3(aq) + Zn(s)
When a material made of iron rusts, its mass increases so does this observations violate the law of conservation of mass?
For the reaction
NH3(g) + H2S(g) → NH4HS(s)
K = 400. at 35.0°C. If 3.50 mol each of NH3, H2S, and NH4HS are placed in a 5.00-L vessel, what mass of NH4HS will be present at equilibrium? What is the pressure of H2S at equilibrium?
mass NH4HS at equilibrium =_________ g
PH2S = ____________atm
The pressure of H2S at equilibrium is 1.50 atm.
The balanced equation for the given reaction is:
NH3(g) + H2S(g) → NH4HS(s)
The equilibrium constant, K, is given as 400 at 35.0°C.
Initially, 3.50 mol each of NH3, H2S, and NH4HS are placed in a 5.00-L vessel.
Let x be the amount of NH4HS that is formed at equilibrium.
The equilibrium concentrations can be calculated using an ICE table:
NH3(g) + H2S(g) ⇌ NH4HS(s)
Initial: 3.50 3.50 3.50
Change: -x -x +x
Equilibrium: 3.50-x 3.50-x 3.50+x
Using the equilibrium concentrations, we can write the expression for the equilibrium constant:
K = [NH4HS]/([NH3][H2S])
Substituting the equilibrium concentrations, we get:
400 = (3.50 + x)/[(3.50 - x)(3.50 - x)]
Simplifying and solving for x, we get:
x = 1.33 mol
Therefore, the amount of NH4HS that is formed at equilibrium is 1.33 mol.
The mass of NH4HS can be calculated using its molar mass, which is:
NH4HS: 53.08 g/mol
Mass of NH4HS = 1.33 mol x 53.08 g/mol = 70.5 g
Therefore, the mass of NH4HS that is present at equilibrium is 70.5 g.
To calculate the pressure of H2S at equilibrium, we need to know the number of moles of H2S present at equilibrium. From the ICE table, we can see that the change in the concentration of H2S is -x, so the number of moles of H2S at equilibrium is:
n(H2S) = 3.50 mol - x = 2.17 mol
The total pressure at equilibrium is the sum of the partial pressures of NH3, H2S, and the vapor pressure of NH4HS, which we assume to be negligible. Since NH3 and H2S are both ideal gases, their partial pressures can be calculated using the ideal gas law:
PV = nRT
For NH3, we have:
P(NH3) = n(NH3)RT/V
Substituting the values, we get:
P(NH3) = (3.50 - x)RT/V = (3.50 - 1.33) mol x 0.0821 L atm mol^-1 K^-1 x (35.0 + 273.15) K / 5.00 L = 2.37 atm
For H2S, we have:
P(H2S) = n(H2S)RT/V
Substituting the values, we get:
P(H2S) = 2.17 mol x 0.0821 L atm mol^-1 K^-1 x (35.0 + 273.15) K / 5.00 L = 1.50 atm
Therefore, the pressure of H2S at equilibrium is 1.50 atm.
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Suppose you want to search for high-redshift star-forming galaxies using a telescope equipped with a spectrograph able to measure the entire optical spectrum (400-700 nm). Star-forming galaxies contain copious hydrogen gas, from which stars form. Some of this gas will be ionized by the newly-formed stars; the spectra of star-forming galaxies there exhibit bright hydrogen lines in emission. Light from the newly-formed stars is absorbed by neutral hydrogen gas as it passes through the galaxy. a) If you wish to search for Lyman-a emitting galaxies, over what redshift range can you find such galaxies? [3 points] b) If you detect only one line, you cannot be certain that this line is the Lyman-a line, and hence that the galaxy is indeed at the computed redshift. Assuming you can also observe in the infrared (wavelengths >700 nm), how can you change your strategy to make sure that the line you detect is really the Lyman-a line? Give three examples of how you can increase confidence in the correct identification of the Lyman-a line considering only hydrogen gas. [3 points] c) Limited only to optical wavelengths and considering only hydrogen gas, what strategy should you adopt to be certain that the line you detect is really the Lyman-a line while maximizing the redshift range over which you find galaxies? With this strategy, over what redshift range can you find star-forming galaxies? Justify through appropriate computations and reasoning that this is in fact the optimal strategy for maximizing the redshift range of your search. [10 points] d) Apart form hydrogen emission lines, star-forming galaxies also usually exhibit bright [OIII] forbidden lines. Why can such lines be seen from interstellar gas but not the Earth's atmosphere or in the laboratory? [4 points]
a) The maximum observable range of redshifts that produces Lyman-alpha line is 0 ≤ z ≤ 10.6
b) i) identifying the galaxy with a radio source, ii) looking for other Lyman lines, iii) a coincidence with a continuum break
c)The maximum redshift range over which galaxies can be found using this strategy is z = 7 to z = 15.5.
d)Earth's atmosphere absorbs the radiation, and the laboratory conditions are not the same as interstellar conditions.
a) Lyman-alpha line is produced by the hydrogen atoms that have electrons that are in the ground state being raised to the first excited state. Over a certain range of redshifts, the Lyman-alpha line is redshifted to longer wavelengths that are observable by an optical spectrograph. The maximum observable range of redshifts that produces Lyman-alpha line is 0 ≤ z ≤ 10.6 (depending on the exact details of the galaxy's emission profile).
b) Observing the galaxy in the infrared can help in the identification of the Lyman-alpha line as it is shifted to longer wavelengths. Three ways to increase confidence in the correct identification of the Lyman-alpha line are:
i) identifying the galaxy with a radio source, ii) looking for other Lyman lines, iii) a coincidence with a continuum break.
c) The strategy that needs to be adopted is to look for the Lyman limit, which is the point at which the spectrum is cut off by the absorption of all hydrogen in the galaxy. To be certain that the line you detect is the Lyman-alpha line, you need to look for a decrement in the flux of the galaxy at wavelengths shorter than the line and a decrement in the flux at wavelengths longer than the line. This is because the Lyman limit will be shifted to longer wavelengths at higher redshifts, so to maximize the redshift range over which galaxies can be found, you need to search for the Lyman limit at the longest wavelength possible. The maximum redshift range over which galaxies can be found using this strategy is z = 7 to z = 15.5.
d) The reason why such lines can be seen from interstellar gas but not the Earth's atmosphere or in the laboratory is that the Earth's atmosphere absorbs the radiation, and the laboratory conditions are not the same as interstellar conditions. The forbidden lines from the interstellar gas are not affected by dust absorption because they are produced in regions where dust is not present.
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Why does every chemical reaction require a certain amount of activation energy?
A Energy is released when the reactants begin to react.
B. Energy lost to the environment during the reaction must be replaced.
C. Forming the activated complex requires energy.
D. The products have more potential energy than the activated complex.
E. The reactants have less potential energy than the products.
Answer: C
Explanation:
What is the condensation AND freezing point of water C AND F!!
Answer:
Condensation 212F or 100C, Freezing 32F or 0C
Explanation:
Condensation 212 degrees Fahrenheit or 100 degrees Celsius.
Freezing point 32 degrees Fahrenheit or 0 degrees Celsius.
What is 0.866 km to mm in scientific notation
Answer:
866000
Explanation:
because one kilometer (km) = 1000000 millimeters (mm)
How many moles are in 18.8 grams of NaOH
A student produced 1.06 g of NH3 from the reaction of 1.77 g of H2 with 25.7g N2 gas.
N2 + 3H2 → 2NH3
Theoretical yield?
Percent yield?
The theoretical yield of NH3 is 10.03 g and percent yield is 10.56%
What is the theoretical yield?The theoretical yield is obtained from the eequation if reaction.
The equation of the reaction is given below:
N2 + 3H2 → 2NH3moles of reactants and products:
moles of H2 = 1.77/2 = 0.885 moles
moles of N2 = 25.7/28 = 0.918 moles
moles of NH3 = 1.06/17 = 0.062
From the equation of the reaction, H2 is the limiting reactant
3 moles of H2 produces 2 moles of NH3
0.885 moles of H2 will produce 2/3 × 0.918 = 0.59 moles of NH3
Therefore:
theoretical yield = 0.59 × 17
theoreticalyield = 10.03 g of NH3
Percent yield = 1.06/10.03 × 100%
percent yield = 10.56%
Therefore, the theoretical yield is 10.03 g and percent yield is 10.56%.
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What is the scientific method and what the second step of the scientific ? If you answer the question right you will get a 5 star and a brainly
Answer:
a method of procedure that has characterized natural science since the 17th century, consisting in systematic observation, measurement, and experiment, and the formulation, testing, and modification of hypotheses.The basic steps of the scientific method are: 1) make an observation that describes a problem, 2) create a hypothesis, 3) test the hypothesis, and 4) draw conclusions and refine the hypothesis.
Explanation:
the scienticif Method and the second step of the scientific
In Niels Bohr’s model of the atom, how are electrons configured?
In Niels Bohr’s model of the atom, electrons are configured in a series of concentric shells around the nucleus. The shells are numbered, with the shell closest to the nucleus being numbered one, and each succeeding shell numbered two, three, and so on.
The electrons in the innermost shell have the lowest energy, while those in the outermost shell have the highest energy. Each shell can hold a certain number of electrons. The first shell can hold up to two electrons, the second shell up to eight electrons, and the third shell up to 18 electrons. Electrons fill the shells in a specific order, following the Aufbau principle. The principle states that electrons will occupy the lowest available energy level before filling higher levels. Electrons in the same shell have the same energy. Electrons in different shells have different amounts of energy, which corresponds to the distance of the shell from the nucleus. When an electron absorbs energy, it can move to a higher energy level. When an electron loses energy, it can move to a lower energy level. Electrons can also move between atoms, which is the basis of chemical reactions.For such more question on concentric shells
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TRUE OR FALSE
- during all chemical reactions, the mass of the products is never equal to the mass of the reactants
- Mass is sometimes lost in chemical reactions
- The Law of Conservation of Mass states that mass is not conserved
During all chemical reactions, the mass of the products is never equal to the mass of the reactants.
- FALSE
Mass is sometimes lost in chemical reactions.
- FALSE
The Law of Conservation of Mass states that mass is not conserved.
- FALSE
Explain how the two atoms in a chlorine molecule are held together
The chlorine atoms can be held together by the dispersion forces.
How are the chlorine atoms held together?We know that the chlorine atoms are the atoms that we could say that they have an electronegativity value that is almost the same. As such we can see that the the compound is non polar.
Given the fact that the compound is non polar, the atoms would be held together by a kind of bond that not polar in nature and these are the dispersion forces that hold the non polar molecules together.
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Calculate the number of moles in butane C4H10 in 151g of butane (at masses c =12 amu and H=1 )
The number of moles in butane C4H10 in 151gm of butane is 2.605 moles.we can do this with the help of molecular weights of C,H.
To calculate the number of moles in 151 grams of butane (C4H10), we need to determine the molecular weight of butane. The molecular weight of butane can be calculated using the atomic masses of its elements:
1 mole of C = 12 g 1 mole of H = 1 g
So, the molecular weight of butane can be calculated as: 4 moles of C * 12 g/mole + 10 moles of H * 1 g/mole = 4 * 12 + 10 * 1 = 48 + 10 = 58 g/mole
Now that we know the molecular weight of butane, we can use it to calculate the number of moles in 151 grams of butane:
151 g of butane / 58 g/mole = 2.605 moles
Therefore, there are 2.605 moles of butane in 151 grams of butane.
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9- what are the stoichiometric coefficients required to balance the following equation? c6h14(g) o2(g) co2(g) h2o(l)
The stoichiometric coefficients required to balance the equation C6H14(g) + O2(g) ⇒CO2(g) + H2O(l) are: 1, 9, 6, 7. The correct option is B.
The equation must be balanced so that the number of atoms of each element is equal on both sides of the equation. To balance this equation, we start by counting the number of atoms of each element on both sides.
On the left-hand side, there are 6 carbon atoms and 14 hydrogen atoms. On the right-hand side, there are 6 carbon atoms and 14 hydrogen atoms in the H2O, and 1 carbon atom and 2 oxygen atoms in the CO2. Therefore, we need to balance the equation by adding coefficients in front of the reactants and products to make sure the number of atoms of each element is equal on both sides.
By adding coefficients, the balanced equation becomes:
C6H14(g) + 9O2(g) ⇒ 6CO2(g) + 7H2O(l)
Therefore, the stoichiometric coefficients required to balance the equation are 1, 9, 6, 7. The correct option is B.
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complete question:
What are the stoichiometric coefficients required to balance the following equation? C6H14(g) + O2(g) ⇒ CO2(g) + H2O(l)
A. 2, 19, 12, 14
B. 1, 9, 6, 7
C. 1, 7, 6, 7
D. 1, 19, 6, 7
5.00 g of glucose, C6H12O6, is dissolved in 72.8 g of water. What will be the freezingpoint and the boiling point?
Answer
The freezing point o the solution = 273.714153 K
The boiling point of the solution= 373.1955328 K
Explanation
Given:
Mass of glucose = 5.00 g
Volume of water = 72.8 g
What to find:
The freezing point and the boiling point of the solution.
Step-by-step solution:
Note: (Freezing point of water = 273K, Kf for water =1.87K kg/mol, atomic weight C = 12, H = 1, O = 16).
The freezing point of the solution:
The molecular weight of glucose C6H12O6 = 6(12) + 12(1) + 6(16) = 180 g/mol
The number of moles of glucose = (Mass of glucose/Molecular weight) = 5.00 g/180.0 g/mol = 0.0278 moles
Mass of water = 72.8 g = 0.0728 kg
So molality of glucose = (Moles o glcsose/Volume of solution) = 0.0278 mol/0.0728 kg = 0.3819 mol/kg
The depression in the freezing point, ∆T = Kf x molality = 1.87 K kg/mol x 0.3819 mol/kg = 0.714153 K
Since the freezing point of water = 273 K
Therefore, the freezing point of the solution= 273 K + 0.714153 K = 273.714153K
The boiling point of the solution:
∆T = i x m x Kb
∆T = change in temperature i.e boiling point elevation
i = van't Hoff factor = 1 for glucose since it does not ionize or dissociate. It is a single particle.
m = molality = moles solut/kg solvent = 0.0278 mol/0.0728 kg = 0.3819 mol/kg
Kb = boiling poin constant = 0.512 K kg/mol
∆T = 1 x 0.3819 mol/kg x 0.512 K kg/mol
∆T = 0.1955328 K
Since the freezing point of water = 373 K
Therefore, the boiling point of the solution = 373 K + 0.1955328 K = 373.1955328 K
2. write a conclusion statement that will address the following:
write in complete sentences.
is your hypothesis supported or not? explain using your data/calculations for support.
the accuracy & precision of your data.
possible further experimentation.
possible sources of error
.
(the hypothesis is : if the density of the object is 19.32 g/ml then the object is gold)
Based on the data and calculations obtained, it can be concluded that the hypothesis stating, "If the density of the object is 19.32 g/ml, then the object is gold" is not supported.
To support this conclusion, the data and calculations regarding the density of the object should be analyzed. If the calculated density of the object differs significantly from the expected value of 19.32 g/ml, it indicates that the object is not gold.
The accuracy and precision of the data can be assessed by comparing the calculated density with the expected density value. If the calculated density is close to the expected value, it suggests high accuracy. Additionally, if multiple measurements of density yield consistent results, it indicates high precision.
Further experimentation could involve additional tests to determine the identity of the object. These tests could include assessing other physical or chemical properties such as melting point, electrical conductivity, or reactivity with certain substances.
Possible sources of error in the experiment could include instrumental errors in measuring the mass or volume of the object, contamination of the object, or inaccuracies in the known density of gold used for comparison. These factors could contribute to deviations between the calculated and expected densities.
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7. What is the mole ratio of NH3 to N₂ in the balanced chemical equation below?
4NH3(g) + 6NO (g) → 5N2 (g) + 6H₂O (l)
A. 3:2
B. 4:5
C. 10:11
D. 1:1
8. What mass of glucose (C6H12O6) can be produced from a photosynthesis reaction that occurs using 10
moles of CO₂?
6CO₂ (g) + 6H₂O (1)→ C6H12O6 (aq) + 602 (g)
A 10 grams
B. 180 grams
C. 300 grams
D. 1,080 grams
9. During photosynthesis, how many moles of water (H₂O) are needed to produce 150 grams of glucose
(C6H12O6)?
A. 2 moles
B. 3 moles
C. 4 moles
D. 5 moles
10. During photosynthesis, how many grams of oxygen gas (0₂) are produced from 55 grams of carbon
dioxide gas (CO₂)?
A. 32 grams 0₂
B. 40 grams 0₂
C. 44 grams 0₂
D. 55 grams O₂
7) From the coefficients of the equation, the answer is 4:5.
8) From the coefficients of the equation, for every 6 moles of carbon dioxide consumed, 1 moleof glucose is consumed. This means that 1.66 moles of glucose is produced. Glucose has a formula mass of 180.16 g/.mol, so the answer is (1.66)(180.16), which is about 300 grams.
9) From the coefficients of the equation, we know that for every 6 moles of water consumed, 1 mole of glucose is consumed. Since we want 150 grams of glucose, and glucose has a formula mass of 180.16 g/mol, we want 150/180.16 = 0.83 mol of glucose. Thus, we need 0.83(6) moles of water, which is 5 moles to 1 sf.
10) From the coefficients of the equation, we know that for every 6 moles of carbon dioxide produced, 6 molds of oxygen are produced. Since carbon dioxide has a formula mass of 44.009 g/mol, 55 grams is about 1.25 moles. Thus, since oxygen has a formula mass of about 32 g/mol, the answer is 32(1.25), which is about 55 grams.
The first part of the strontium test removes any residual barium. Do you have to be careful adding too much additional chromate? What might happen to the strontium ?
Yes, it is necessary to be careful when adding too much additional chromate during the strontium test. Excessive amounts of chromate can form a precipitate with strontium ions, leading to the formation of strontium chromate.
This can interfere with the accurate detection and measurement of strontium. Strontium chromate is a yellow solid that can precipitate out of the solution, making it difficult to distinguish and quantify the presence of strontium. This interferes with the accuracy and reliability of the strontium test. Therefore, it is important to use the appropriate amount of chromate in the test to ensure that the reaction specifically targets the barium ions without affecting the strontium ions.
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Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.
Answer:
See Explanation
Explanation:
The question is incomplete; as the mixtures are not given.
However, I'll give a general explanation on how to go about it and I'll also give an example.
The percentage of a component in a mixture is calculated as:
\(\%C_E = \frac{E}{T} * 100\%\)
Where
E = Amount of element/component
T = Amount of all elements/components
Take for instance:
In \((Ca(OH)_2)\)
The amount of all elements is: (i.e formula mass of \((Ca(OH)_2)\))
\(T = 1 * Ca + 2 * H + 2 * O\)
\(T = 1 * 40 + 2 * 1 + 2 * 16\)
\(T = 74\)
The amount of calcium is: (i.e formula mass of calcium)
\(E = 1 * Ca\)
\(E = 1 * 40\)
\(E = 40\)
So, the percentage component of calcium is:
\(\%C_E = \frac{E}{T} * 100\%\)
\(\%C_E = \frac{40}{74} * 100\%\)
\(\%C_E = \frac{4000}{74}\%\)
\(\%C_E = 54.05\%\)
The amount of hydrogen is:
\(E = 2 * H\)
\(E = 2 * 1\)
\(E = 2\)
So, the percentage component of hydrogen is:
\(\%C_E = \frac{E}{T} * 100\%\)
\(\%C_E = \frac{2}{74} * 100\%\)
\(\%C_E = \frac{200}{74}\%\)
\(\%C_E = 2.70\%\)
Similarly, for oxygen:
The amount of oxygen is:
\(E = 2 * O\)
\(E = 2 * 16\)
\(E = 32\)
So, the percentage component of oxygen is:
\(\%C_E = \frac{E}{T} * 100\%\)
\(\%C_E = \frac{32}{74} * 100\%\)
\(\%C_E = \frac{3200}{74}\%\)
\(\%C_E = 43.24\%\)
if you have an atom from group 2 of the periodic table, and it reacts with an atom from group 17 of the periodic table, what is the general formula?
When an atom from group 2 of the periodic table(say X) reacts with an atom of group 17(say Y), then the general formula is XY₂.
The elements present in group 2 of the periodic table (Be, Mg, Ca, Sr, Ba, Ra) are metals having valency 2. On the other hand, the elements present in group 17 ( F, Cl, Br, I, At, Ts) are halogens or non-metals having valency 1. So, elements from both these groups combine together by transfer of electrons, forming an ionic bond between them.
The general formula is formed by exchange of valencies.
here, valency of X=2, Y=1
Thus, the general formula is XY₂
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explain how components of the atmosphere can be used successfully in producing important chemicals
Oxygen and nitrogen from the atmosphere can be used as feedstock to produce chemicals such as ammonia, nitric acid, and sulfuric acid.
What are chemicals?The components of the atmosphere, such as nitrogen and oxygen, can be used to produce important chemicals through industrial processes such as the Haber-Bosch process for ammonia synthesis and the production of nitric acid. Nitrogen and oxygen can also be used as oxidizers in combustion processes to produce energy and heat, such as in the burning of fossil fuels.
Additionally, carbon dioxide from the atmosphere can be used as a feedstock for the production of chemicals such as methanol and formic acid through processes like carbon capture and utilization. The use of atmospheric components in chemical production can help to reduce reliance on non-renewable resources and support the development of sustainable manufacturing processes.
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Task 3: Below are data you collected from a
reservoir. Complete tasks 3a and 3b.
Hardap Dam water parameters for different depths
taken:
Depth
Oxygen
pH
Temperature
Surface
7.5
7.9
It can be assumed that it is within the acceptable range for domestic use. Rapid temperature changes or high temperatures can also cause damage to pipes and fittings.
Task 3a: Discuss the significance of the measured water parameters collected from Hardap Dam for aquatic life.Water parameters are chemical, biological and physical characteristics of the water. They are important indicators of the quality of water. Hardap Dam is a habitat for a variety of aquatic life such as fish, birds, plants and insects. The measured water parameters for different depths include oxygen, pH and temperature.The concentration of oxygen in the water is crucial to aquatic life. Oxygen is required for respiration by aquatic animals. The surface oxygen concentration of 7.5 mg/L measured in Hardap Dam is adequate for most aquatic life. However, some fish species require higher concentrations of dissolved oxygen to survive. As water depth increases, the oxygen concentration decreases.
This can be seen in the decreasing oxygen concentration at depths below the surface. Low oxygen concentration can lead to suffocation of aquatic life, changes in species composition and nutrient cycling.PH is a measure of the acidity or alkalinity of water. Aquatic life requires a pH range of 6.5-9.0 to survive. The pH of 7.9 measured at the surface of Hardap Dam indicates that the water is slightly alkaline. The pH values for the deeper water layers were not provided but it can be assumed that they are likely to be similar. Extreme pH values can lead to stress and death of aquatic organisms.Temperature is an important parameter that influences the metabolic rates of aquatic organisms. Temperature affects the solubility of oxygen and other gases in water. It also determines the rate of biochemical reactions in organisms. The temperature of the water at the surface of Hardap Dam was not provided.
However, it can be assumed that it is within the range of tolerance for most aquatic organisms.
As water depth increases, temperature decreases. Rapid temperature changes or high temperatures can cause stress and death in aquatic organisms. Task 3b: Evaluate the significance of the measured water parameters in relation to the use of Hardap Dam water for domestic purposes.Water is a critical resource for human survival. Hardap Dam supplies water to communities for domestic purposes. The water quality is important to prevent the spread of diseases and illness. The measured water parameters for different depths of the dam include oxygen, pH and temperature.
These parameters affect the suitability of the water for domestic use.Oxygen concentration in the water is important for the removal of pollutants and odours. The surface oxygen concentration of 7.5 mg/L measured in Hardap Dam is adequate for this purpose. As the water depth increases, oxygen concentration decreases. Low oxygen concentration can lead to unpleasant tastes and odours in the water.PH is important for the taste and aesthetics of water. The pH of 7.9 measured at the surface of Hardap Dam is within the acceptable range for domestic use. Extreme pH values can cause water to taste bitter or metallic.
Changes in pH can also affect the corrosion of pipes and fittings.Temperature can affect the growth of microorganisms in the water. High temperatures can promote the growth of harmful bacteria such as E. coli. The temperature of the water at the surface of Hardap Dam was not provided. However, it can be assumed that it is within the acceptable range for domestic use. Rapid temperature changes or high temperatures can also cause damage to pipes and fittings.
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Radon-22 has a half –life of approximately 6 days. If you begin with a sample containing 40 grams of radon, how many grams of radon will be left after 24 days?
On the IR spectrum of Adipic Acid, which you took in lab, indicate where each of the following significant bonds stretches appear on the IR and submit this IR with your report. Carboxylic acids: Co, C-o. o-H Alkyl group/chain: sp'c-H
The IR spectrum of Adipic Acid will show the following absorption bands for the specified bonds:
Carboxylic acids: The C=O bond (carbonyl) stretch will appear at around 1700-1710 cm⁻¹. The C-O stretch will appear at around 1300-1350 cm⁻¹.o-H: The O-H stretch will appear at around 3400-3500 cm⁻¹.Alkyl group/chain: The sp3-C-H stretch will appear at around 2900-3000 cm⁻¹. IR spectrum of Adipic AcidThe IR spectra of a compound can be used to identify the functional groups present in the compound by looking for specific absorption bands that correspond to the vibrations of the bonds in those functional groups.
The C=O bond stretch, also called carbonyl, is a characteristic absorption of carboxylic acids and it appears at around 1700-1710 cm⁻¹.The C-O stretch is also a characteristic absorption of carboxylic acids and it appears at around 1300-1350 cm⁻¹.The O-H stretch is a characteristic absorption of hydroxyl groups, it appears at around 3400-3500 cm⁻¹.The sp3-C-H stretch is a characteristic absorption of alkyl groups and it appears at around 2900-3000 cm⁻¹.To confirm the presence of these functional groups, we can compare the IR spectrum of the compound with known spectra of compounds that contain these functional groups. Additionally, the chemical shifts and relative intensities of the peaks can also be used to confirm the presence of these functional groups.
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part a when a ketone and its enol are in equilibrium, under most conditions the concentration of the enol is ________ the concentration of the ketone.
Answer:
Much lower than ketone is more stable than enol. N, 4-Dimethylpent -4-en-2-Amine (NH_3 protonated in acidic protoned in acidic conditions) d. [Proton cannot re extracted from OH in acidic conditions to firm O^(-)]
-. A 100.0 g Chunk of Aluminum with an Initial Temperature of 450.0 C° is added to 100.0
mL of Ethyl Alcohol with an Initial Temperature of 80.0 C° (A) Calculate Equilibrium
Temperature of the Mixture (B) Calculate the Heat Exchange of the System
BA) 198 C° B) 22770 J All answers Approx
D A) 110C° B) 10000 J All answers Approx
A) 150 C° B) 30000 J All answers Approx
A) 300 C° B) 15000 J All answers Approx
The negative sign means that heat was transferred from the aluminium piece to the ethyl alcohol. Consequently, the heat exchange of the system is roughly -22,770 J, or -2.28 x 10^4 J (to 2 significant figures).
What happens when 100g of 100 C boiling water is introduced to a calorimeter?The mixture's temperature rises to 20 C. The mixture in the calorimeter is then dipped into by a metallic block of mass 1 kilogramme at 10°C. The temperature rises to 19 C once thermal equilibrium has been reached.
Q = m * c * T, where Q is the heat exchanged, m is the mass of the substance, c is the specific heat capacity, and T is the change in temperature, is the formula used to determine the heat exchanged.
We can use the following formula to determine the equilibrium temperature:
m_al * c_al * (T_eq - T_al) = m_et * c_et * (T_et - T_eq)
By entering the specified values, we obtain:
(0.100 kg) * (0.902 J/g°C) * (T_eq - 450.0°C) = (0.100 kg) * (2.44 J/g°C) * (80.0°C - T_eq)
Simplifying, we get:
90.2 J/C * (T_eq - 450.0) = 244 J/C * (80.0 - T_eq)
90.2 T_eq - 40590 = 19520 - 244 T_eq
334.2 T_eq = 60110
T_eq ≈ 179.7°C ≈ 180°C (to the nearest 10°C)
As a result, the mixture's equilibrium temperature is roughly 180 °C.
(B) We can apply the same formula as before to determine the system's heat exchange. We can suppose that the mixture has a specific heat capacity equal to that of ethyl alcohol.
Q = m_al * c_al * (T_eq - T_al) + m_et * c_et * (T_eq - T_et)
Plugging in the given values, we get:
Q = (0.100 kg) * (0.902 J/g°C) * (180.0°C - 450.0°C) + (0.100 kg) * (2.44 J/g°C) * (180.0°C - 80.0°C)
Q ≈ -22,770 J ≈ -2.28 x 10⁴ J (to 2 significant figures)
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7. Potassium chlorate is often used for pyrotechnics and fireworks because it gives off so
much heat and light when it decomposes. How much heat will 253 grams of potassium
chlorate release if the standard enthalpy change for this reaction is -91 kJ?
Enthalpy, in a technical sense, refers to the internal energy needed to create a system as well as the energy needed to create space for it by establishing its pressure, volume, and displacing its surroundings.
In a thermodynamic system, energy is measured by enthalpy. Enthalpy is a measure of a system's overall heat content and is equal to the system's internal energy plus the sum of its volume and pressure.
A state function that is entirely based on state functions P, T, and U is how enthalpy is also described.
Here the equation used is:
q = n × ΔH
n = Mass / Molar mass
n = 253 / 122.55 = 2.064 mol
q = 2.064 × -91 = -187.82 kJ
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(a) what is the total random kinetic energy of all the molecules in one mole of hydrogen at a temperature of 300 k? (b) with what speed would a mole of hydrogen have to move so that the kinetic energy of the mass as a whole would be equal to the total random kinetic energy of its molecules?
Total random kinetic energy of all the molecules in one mole of hydrogen at temperature of 300 K is 3741.3 J. Speed with which a mole of hydrogen have to move so that kinetic energy of the mass is equal to the total random kinetic energy of its molecules is 7482.6 m/s
What is kinetic energy of gas molecule?According to kinetic theory, average kinetic energy of gas molecules depends on the absolute temperature. At given temperature, molecules of all the gases have same average kinetic energy".
Kinetic energy = 3/2 * R T
= 3/2 * 8.314 * 300
= 3741.3 J
KE = 1/2 * m v²
v² = 2 KE/m
= 2* 3741.3 /1
Speed = 7482.6 m/s
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What is the naming process of alkanes, alkenes, and alkynes
Answer: Heyo Kenji Here! Here's your answer- Higher alkenes and alkynes are named by counting the number of carbons in the longest continuous chain that includes the double or triple bond and appending an -ene (alkene) or -yne (alkyne) suffix to the stem name of the unbranched alkane having that number of carbons.
Explanation: Hope this helps!
Have a nice day!
- Kenji ^^
Answer:
Alkene and alkyne compounds are named by identifying the longest carbon chain that contains both carbons of the double or triple bond.
Explanation:
How to separate pollen from air
Answer:
by removibg it
Explanation: