[tex]\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}[/tex]
dt remains same for reaction[tex]\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}[/tex]
[tex]\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})[/tex]
[tex]\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}[/tex]
M is molarity here not metre
The average rate of appearance of H2 over the time period from t = 0 s to [tex]\rm t = 5.48\times10^3 s\ is\ 2.25\times10^{-6}[/tex] M/s.
To find the average rate of appearance of [tex]\rm H_2[/tex], we can use the stoichiometry of the reaction to relate it to the rate of disappearance of [tex]\rm NH_3[/tex].
From the balanced equation: [tex]\rm 2NH_3(g) - > N_2(g) + 3 H_2(g)[/tex]
We can see that for every 2 moles of [tex]\rm NH_3[/tex] that disappear, 3 moles of [tex]\rm H_2[/tex] appear.
Given that the average rate of disappearance of [tex]\rm NH_3\ is\ 1.50\times10^{-6} M/s[/tex], we can calculate the average rate of appearance of [tex]\rm H_2[/tex] as follows:
Average rate of appearance of [tex]\rm H_2[/tex] = (3/2) x (Average rate of disappearance of [tex]\rm NH_3[/tex])
Average rate of appearance of [tex]\rm H_2[/tex] = (3/2) x [tex]\rm (1.50\times10^{-6} M/s)[/tex]
Average rate of appearance of [tex]\rm H_2[/tex] = [tex]\rm 2.25\times10^{-6} M/s[/tex]
Therefore, the average rate of appearance of [tex]\rm H_2[/tex] over the time period from t = 0 s to [tex]\rm t = 5.48\times10^3 s\ is\ 2.25\times10^{-6}[/tex] M/s.
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What volume of a 3.50 M Na3PO4 solution would you use to make 1.50 L of a 2.55 M Na3PO4 solution?
A. 1.09 L
B. 2.06 L
C. 13.4 L
D. 0.917 L
why neutron doesn't contain any charge?
Answer:
Neutron does not contain any charge because the charge of the quarks that made up the neutron balances each other out.
Hope it helps.
In the second equation FePO4 is an ionic compound. What is the charge on the Fe?
Answer:
+3
Explanation:
The phosphate ion has a charge of -3, so for the compound to have an overall change of 0, Fe needs to have a charge of +3.
Oil from cashew nuts
Answer:
yes it's a thing, milk too
For the reaction: C → D + B the graph of ln k vs 1/T(K) gives the equation: y = -2.2x104x + 45.0. Using this information, calculate the activation energy, (J/mol)
From the equation of the graph, the activation energy of the reaction is -2.2 * 10⁴ J/mol.
What is activation energy?Activation energy is the minimum required energy that reactant molecules must possess for a reaction to proceed towards formation of products.
The activation energy is determined from the slope of ln k against 1/T(K).
Given the equation, y = -2.2 * 10⁴ x + 45.0.
Comparing with the equation of a straight line, y = mx + c
The gradient, m = -2.2 * 10⁴
Therefore, the activation energy of the reaction is -2.2 * 10⁴ J/mol.
In conclusion, reactant molecules must break the activation energy barrier in order to form products.
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James was supposed to measure the circumference of three trees as homework for science class. When he got to class the next day, he realized that he measured the trees in inches instead of centimeters.
One tree had a circumference of 41 inches. What is the tree's circumference in centimeters?
Hint: 1 inch equals 2.5 cm
A.
102.5 cm
B.
27 cm
C.
135 cm
D.
16 cm
Answer:
[tex]\huge\boxed{\sf 102.5\ cm}[/tex]
Explanation:
Given that,
Circumference = C = 41 inches
Also,
1 inch = 2.5 cmMultiply 41 to both sides
41 inches = 2.5 × 41 cm
41 inches = 102.5 cm
So, The tree has a circumference of 102.5 cm.
[tex]\rule[225]{225}{2}[/tex]
The half-life of the radioactive isotope magnesium-20 is 0.600 seconds.
How long will it take for the mass of a sample of magnesium-20 to decay from 65.6 micrograms to 8.20 micrograms?
____________seconds
It will take less 1.8 seconds for the mass of a sample of magnesium-20 to decay from 65.6 micrograms to 8.20 micrograms.
What is Half-Life of radioactive a radioactive substance?The half-life of a radioactive element is the time for half the amount of a sample of the substance to decay.
After 0.6 seconds 31.25 remains
After 1.2 seconds, 15.625 remains
After 1.8 seconds, 7.9 micrograms remains.
In conclusion, it will take less 1.8 seconds for the sample to decay to 8.20 micrograms.
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The time taken for the radioactive sample to decay is 1.8 s.
What is the half life?The half life is the time taken for only half of the number of the radioactive isotopes to remain. Now we have;
No = 65.6 micrograms
N = 8.20 micrograms
t1/2 = 0.600 seconds
t = ?
Hence;
N/No = (1/2)^t/t1/2
8.20/ 65.6 = (1/2)^t/0.600
0.125 = (1/2)^t/0.600
1/8 = (1/2)^t/0.600
(1/2)^3 = (1/2)^t/0.600
3 = t/0.600
t = 3 * 0.600
t = 1.8 s
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what is entropy?
Answer this and Have a good day
Answer:
Entropy is a measurement of how much the atoms in a substance are free to spread out, move around, and arrange themselves in random ways
21
Select the correct answer.
How many valence electrons does oxygen have?
OA 2
OB. 4
О с. 6
OD. 8
OE. 10
Answer: 6 valence electrons
Explanation: the atomic number for oxygen is 8. the first shell takes 2, the second 8. so, the outer shell is the last shell, which takes 6 from oxygen because oxygen has only 8. 6 is the number for outer or valence electron for oxygen
A solution is prepared by dissolving 0.131 g of a substance in 25.4 g of water. The molality of the solution is determined by freezing point
depression to be 0.056 m. What are the moles of the substance?
The mole of the substance, given the data from the question is 0.0014 mole
What is molality?This is simply defined as the mole of solute per kilogram of water. Mathematically, it is expressed as
Molality = mole / mass (Kg) of water
How to determine the mole of the substanceMass of water = 25.4 g = 25.4 / 1000 = 0.0254 KgMolality = 0.056 mMole of substance =?Mole = molality × mass of water
Mole of substance = 0.056 × 0.0254
Mole of substance = 0.0014 mole
Thus, the mole of the substance is 0.0014 mole
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What is the density of a liquid with a mass of 6 grams, and a volume
of 3 milliliters?
Answer:
2 g/mL
Explanation:
Density can be calculated using the ratio:
Density = mass (g) / volume (mL)
Since you have been given both the mass and volume of the liquid, you can calculate the density.
Mass = 6 grams (g)
Volume = 3 milliliters (mL)
Density = mass / volume
Density = 6 g / 3 mL
Density = 2 g/mL
Use the ideas of the Kinetic Theory to explain the following. A sealed container contains nitrogen gas. The pressure of a gas is due to the molecules of the gas hitting the walls of the container. Explain why the pressure inside the container increases when the temperature is increased.
Answer:
When the temperature increases, the molecules of the gas gain energy. Therefore, they move faster.
This causes the molecules to hit the walls of the container more frequently and with greater force. Hence the pressure inside the container increases.
Calculate the density of a liquid if 25.7g sample has a volume of 22.6mL
[tex]density = \frac{mass}{volume} [/tex]
[tex]d = \frac{25.7}{22.6} = 1.13717 \: g/ml[/tex]
[tex]pay \: attention \: to \: the \: unit \: used \: to \\ express \: the \: density \\ since \: the \: mass \: is \: given \: in \: grams \\ and \: volume \: in \: milliliter \: then \\ the \: density \: is \: expressed \: in \\ grams \: per \: \: milliliter[/tex]
2. A sample of gas has a pressure of 1.4 atm and a volume of 500 mL. If the volume is decreased to 250 mL, what is the new pressure?
Answer: The new pressure is 2.8 atm.
Explanation:
Since our given information only involves pressure and volume, we will use Boyle's Law: [tex]P_{1} V_{1} =P_{2} V_{2}[/tex]. Let's input the data into the formula.
[tex]1.4atm*500mL=P_{2} *250mL[/tex]
We have to rearrange our data a bit now because the volume is supposed to be in liters. To convert from milliliters(mL) to liters(L), you must divide the number by 1000. That is what we will do with both of our given volumes.
[tex]\frac{500mL}{1000} =0.5L[/tex] [tex]\frac{250mL}{1000} =0.25L[/tex]
Now that our volumes are in liters, we can correctly plug in our data.
[tex]1.4atm*0.5L=P_{2} *0.25L[/tex]
Since we do not know the value [tex]P_{2}[/tex], we must rearrange the formula into a stoichiometric equation to find out the answer.
[tex]P_{2}=\frac{1.4atm*0.5L}{0.25L}[/tex] [tex]P_{2}=2.8 atm[/tex]
Our final answer is 2.8 atmospheres.
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If we want to study rest-frame visible light emitted by stars in distant galaxies at a high redshift of z 3, what portion of the electromagnetic spectrum do we need to observe? (hint JWST is meant to study distant galaxies)
The portion of the electromagnetic spectrum do we need to observe is the red part of the This is further explained below.
What is electromagnetic spectrum?Generally, Spectrum refers to the range of wavelengths or frequencies that electromagnetic radiation may cover.
In conclusion, We need to keep an eye on the red part of the electromagnetic spectrum since a large positive redshift causes an increase in wavelength and a corresponding fall in frequency and photon energy (such as light).
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A) If Kb for NX3 is 4.5×10^−6, what is the pOH of a 0.175 M aqueous solution of NX3 ?
B) If Kb for NX3 is 4.5×10^−6, what is the percent ionization of a 0.325 M aqueous solution of NX3 ?
Please show your work so I can understand how to solve similar problems on my own! Thank you!
3.08 is the pOH of a 0.175 M aqueous solution of [tex]NX_3[/tex].
0.215% is the per cent ionization of a 0.325 M aqueous solution of [tex]NX_3[/tex]
What is pH?pH is a measure of how acidic/basic water is.
A)
[tex]NX_3 + H_2O[/tex] →[tex]NHX_3^+ + OH^-[/tex]
Kb = 4.5 x[tex]10^-6[/tex]
Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).
concentration of (NH₄⁺) = concentration of (OH⁻) = x.
x² = Kb x concentration of (NH₃)
x² = 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.
x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)
= 8.367 × 10⁻⁴
pOH = -log(c(OH⁻))
=- log ( 8.367 × 10⁻⁴)
= 3.08
B)
Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.
Concentration of (NX₃) = 0.325 M.
Kb = 4.5 x 10⁻⁶.
[NX₃H⁺] = [OH⁻] = x.
[NX₃] = 0.325 M - x.
Kb = [NX₃H⁺] x [OH⁻] ÷ [NX₃].
4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).
x = 0.0007 M.
Per cent of ionization:
α = 0. 0007 M ÷ 0. 325 M x 100%
= 0.215%.
Hence,
3.08 is the pOH of a 0.175 M aqueous solution of [tex]NX_3[/tex].
0.215% is the per cent ionization of a 0.325 M aqueous solution of [tex]NX_3[/tex]
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Here is the picture for the q plss help !!
When 400g each of water, oil, and sand are taken and heated from room temperature to 700⁰C on identical Bunsen burners. In this case, Substances that take more time to get heated will take more time to cool Hence, Option (A) is correct.
What is Heat transfer ?
Heat transfer is a form of energy transfer and can occur by conduction, convection, and/or radiation.
Heat transfer occurs any time there is a temperature difference between two objects and occurs in the direction of decreasing temperature, meaning from a hot object to a cold object.
Tests show that the transferred heat relies upon three factors—the adjustment of temperature, the mass of the framework, and the substance and period of the substance.
One of the significant impacts of heat transfer is temperature change: warming builds the temperature while cooling diminishes it.
We expect no stage change and no work is done on or by the framework.
In this case, Substances that take more time to get heated will take more time to cool Hence, Option (A) is correct.
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Sucrose decomposes to fructose and glucose in acid solution. When ln [sucrose] is
plotted vs. time, a straight line with slope of -0.208 hr-1 results. What is the rate law for
the reaction?
A) Rate = 0.208 hr-1 [sucrose]2
B) Rate = 0.208 hr-1 [sucrose]
C) Rate = 0.0433 hr [sucrose]2
D) Rate = 0.0433 hr [sucrose]
E) Rate = 0.208 mol L-1hr-1 [sucrose]0
The correct rate law for the reaction is given as Rate = -0.208 hr-1 [sucrose].
What is a first order reaction?A first order reaction is one in which a plot of the logarithm of the concentration against time gives a straight line graph. Now we know that this is a decomposition reaction thus the slope is negative.
Thus, in this case, the correct rate law for the reaction is given as Rate = -0.208 hr-1 [sucrose].
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the energy of the H-H bond is 432 kJ/mol. The energy of the 0=0 is 498 kJ/mol. witch bond is weakest
The weaker bond is the bond between the two hydrogen atoms since it has a lower bond energy.
What is bond energy?The term bond energy has to do with the energy that is required to break a bond. It is also the energy that must be supplied when a bond is formed.
The lower the bond energy of bond, the weaker the bond. Hence, it follows that the weaker bond here is the bond between the two hydrogen atoms since it has a lower bond energy.
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explain how adding soap to water affects the surface tension of water.
Answer:
The surface tension of water is high due to the dipole nature of water, meaning that the water molecules pull on each other by electrical charge. Soap and detergents interrupt this layer, reducing the surface tension.
A visual, though not scientifically accurate, would be a set of magnetic marbles that form a sheet. If you randomly stick in glass marbles the overall strength decreases since they are not all pulling together in order.
There are some substances, called super wetters, generally silicone surfactants that give a truly amazing drop in surface tension, where a small puddle of water can spread to many times it’s own size. These types of surfactants are used in laundry detergent and french fry oil to prevent foaming and in polymers to promote flow.
Hope it helps!
Please mark me as brainlist.
Answer:
The surface tension of water is high due to the dipole nature of water, meaning that the water molecules pull on each other by electrical charge. Soap and detergents interrupt this layer, reducing the surface tension.
A visual, though not scientifically accurate, would be a set of magnetic marbles that form a sheet. If you randomly stick in glass marbles the overall strength decreases since they are not all pulling together in order.
There are some substances, called super wetters, generally silicone surfactants that give a truly amazing drop in surface tension, where a small puddle of water can spread to many times it’s own size. These types of surfactants are used in laundry detergent and french fry oil to prevent foaming and in polymers to promote flow.
Explanation:
Determine from the balanced chemical reaction if the derived moles ratio is correct or incorrect.
The molar ratio of (6 moles of NaBr)/(1 mole of Fe₂S₃) is correct based on the balanced chemical equation.
Balanced chemical reactionThe balanced chemical reaction is given as;
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
From the balanced chemical reaction above, 6 moles of NaBr will also correspond to 1 mole of Fe₂S₃.
Thus, the molar ratio of (6 moles of NaBr)/(1 mole of Fe₂S₃) is correct based on the balanced chemical equation.
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Identify the following changes as physical or chemical. A loaf of risen but unbaked bread. Photo by Elinor D Bread dough rising: physical chemical
The answers include:
A loaf of risen but unbaked bread - chemical change.Photo by Elinor D - chemical change.Bread dough rising - chemical change.What is a Chemical change?
This involves the formation of a new products from substances. In this scenario, a rising bread contains alcohol which evaporates.
Photographs also fall under this category and is therefore an irreversible chemical change.
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The rate constant, k, for a reaction is 0.0354 sec1 at 40°C. Calculate the rate constant for the
same reaction at 125°C if the activation energy is 26.5 kJ/mol.
Answer:
The rate constant of the reaction at 125˚ is [tex]0.3115 \ \text{sec}^{-1}[/tex].
Explanation:
The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as
[tex]k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}[/tex]
[tex]\ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}[/tex]
where [tex]k[/tex] is the rate constant, [tex]E_{a}[/tex] represents the activation energy of the chemical reaction, [tex]R[/tex] is the gas constant, [tex]T[/tex] is the temperature, and [tex]A[/tex] is the frequency factor.
The frequency factor, [tex]A[/tex], is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.
Consider that the rate constant be [tex]k_{1}[/tex] at an initial temperature [tex]T_{1}[/tex] and the rate constant [tex]k_{2}[/tex] at a final temperature [tex]T_{2}[/tex], thus
[tex]\ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)[/tex]
[tex]\rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}[/tex]
Given that [tex]E_{a} \ = \ 26.5 \ \ \text{kJ/mol}[/tex], [tex]R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}[/tex], [tex]T_{1} \ = \ \left(40 \ + \ 273\right) \ K[/tex], [tex]T_{2} \ = \ \left(125 \ + \ 273\right) \ K[/tex], and [tex]k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}[/tex], therefore,
[tex]k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}[/tex]
Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.
SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)
Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.
1) The theoretical yield of SnS2 will be 4.20 grams
2) The percent yield will be 7.93%
Percent yieldFrom the equation of the reaction:
[tex]SnBr_4(aq)+2Na_2S(aq)-- > 4NaBr(aq)+SnS_2(s)[/tex]
The mole ratio of the reactant is 1:2.
Mole of 48.1 mL, 0.478 M SnBr4 = 0.478 x 48.1/100 = 0.023 mols
Mole of 48.8 mL, 0.160 M Na2S = 0.160 x 48.8/1000 = 0.0078 moles
Thus, Na2S is in excess while SnBr4 is limiting.
Mole ratio of SnBr4 and SnS2 = 1:1
Equivalent mole of SnS2 = 0.023 moles
Mass of 0.023 noles SnS2 = 0.023 x 182.81 = 4.20 grams
2) With 0.0333 g of SnS2 recovered, percent yield = 0.333/4.2 x 100 = 7.93%
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Here is the complete question:
Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.
SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)
Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.
1) Calculate the theoretical yield of SnS2. ;
2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.
Which of the following statements is most accurate about how individual atoms produce light?
Group of answer choices
Electrons can spontaneously emit a photon of light without dropping to lower energy levels.
Light is not created by atoms, it just passes through them.
Photons are emitted when an electron jumps to higher energy levels in an atom.
An electron in an excited energy level of an atom will always drop to lower energy levels, emitting very specific wavelengths of light in the process.
Answer:
An electron in an excited energy level of an atom will always drop to lower energy levels, emitting very specific wavelengths of light in the process.
A gas expands and does PV work on its surroundings equal to 322 J. At the same time, it absorbs 132 J of heat from the surroundings. Calculate the change in energy of the gas. Note: PV work means work done by a changing volume against constant pressure. Enter your answer in scientific notation.
From the calculations, the change in energy is - 190 J.
What is the first law of thermodynamics?From the first law of thermodynamics, the energy is neither created nor destroyed but is transformed from one form to another.
From the law;
U = q + w
U = internal energy
q = heat
w = work
Since work is done on the surroundings and the gas absorbs heat then;
U = 132 J - 322 J
U = - 190 J
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Use the information and examples provided in the exploration to determine the maximum (theoretical) amount of CaCO3 in grams, that can be produced from the precipitation reaction
The maximum amount of CaCO3 we can expect is 0.0180 mole x 100 g/mole = 1.80 g.
Calculations and ParametersGiven that:
There would be a precipitate of calcium carbonate from the reaction between sodium carbonate and calcium chloride.
The reaction is:
Na2CO3 (aq) + CaCl2(aq) → CaCO3 (s) + 2 NaCl (aq)
2.00 g CaCl2 x 1 mole/ 111 g CaCl2
= 0.0180 mole CaCl2
Mix it with 2.00 g Na2CO3 x 1 mole/ 106 g
= 0.0189 mole Na2CO3
The maximum amount of CaCO3 we can expect is 0.0180-mole x 100 g/mole = 1.80 g
Therefore, the 1.80 g is the theoretical (calculated) yield of CaCO3 in this example.
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PLEASE HELP!!!!!!!!
Calculate the pH for the following weak acid.
A solution of HCOOH has 0.19M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4 . What is the pH of this solution at equilibrium?
Answer: 2.23
Explanation:
The dissociation of acetic acid is as follows:
HCOOH(aq) ⇋ HCOO− (aq) + H+(aq)
The ICE table for the concentrations of ions is given below. From the table, the concentration of HCOO- and H+ can be found out.
HCOOH → HCOO− + H+
Initial Concentration 0.019 M 0 0
Equilibrium Concentration (0.019−x) M x x
Where,
x is the concentration of the ions at equilibrium.At equilibrium, dissociation constant can be calculated as follows.
[tex]K_{\mathrm{a}}=\frac{x^{2}}{(0.19-x) \mathrm{M}}[/tex]
At equilibrium, the concentration of x is negligible as compared to that of HCOOH.
Substitute the value of Ka in the above equation.
[tex]\begin{aligned}K_{\mathrm{a}} &=\frac{x^{2}}{0.19-x} \\x &=\sqrt{1.8 \times 10^{-4} \times 0.19} \\&=0.00584 \mathrm{M}\end{aligned}[/tex]
Here, the concentration of hydrogen ion is obtained. From the hydrogen ion concentration, the pH of the solution is found out as follows:
[tex]\begin{aligned}\mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\&=-\log (0.00584 \mathrm{M}) \\&=2.23\end{aligned}[/tex]
Therefore, the pH of 0.19M HCCOH is 2.23
A sample of gas has temperature of 372 K and pressure of 3.8 ATM. If the temperature is increased to 550 K, what is the new pressure?
Answer: The correct answer is 5.6atm.
Explanation: Since we only have information on the pressure and temperature, we will use Gay-Lussac's Law: [tex]\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }[/tex]. Now that we have the formula, let's plug in our given information!
[tex]\frac{3.8atm}{372K} =\frac{P_{2} }{550K}[/tex]
We must find the value [tex]P_{2}[/tex]. To do this, we will write a stoichiometric equation to solve this.
[tex]P_{2} =\frac{3.8atm*550K}{372K}[/tex]
Let's input the numbers into a scientific calculator to get our answer!
[tex]P_{2} =\frac{3.8atm*550K}{372K}=5.6 atm[/tex]
Given all of this work, the value [tex]P_{2}[/tex] equals 5.6 atm.
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A red line is observed at 656.3 nm in the spectrum of atomic hydrogen. determine the values of ???? for the beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line.
The beginning and ending energy levels of the electron during the emission of energy that leads to spectral line n(i)= 2 and n(f)=2
calculation :-(i) Calculation of energy of photon:-wavelength(λ) = 656.3 nm = 656.3 x 10^-9
E= hc / λ
E = energy of the photon
h=Planck's constant
c=speed of light
λ= wavelength of the photon
E= (6.626 x 10^-34)(3 x 10^8)/( 656.3 x 10^-9)
E = 3.028 x 10^-37J
(ii) Beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line.It's a red line which means it falls in a visible region which is known as the Balmer series so;
n (f) = 2
ΔE= RH (1/n(f)^2 - 1/n(i)^2)
3.028 x 10^-37 =-2.18 x 10^-18(1/2^2 - 1/n(i))
n(i)= 2
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