Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? The vapor pressure of the water decreases upon addition of the solute. The osmotic properties of the system lead to this behavior. The overall enthalpy of the system decreases upon addition of the solute. The overall entropy of the system increases upon dissolution of this strong electrolyte. The overall enthalpy of the system increases upon dissolution of this strong electrolyte.

Answers

Answer 1

Answer:

Answer: Option (c) is the correct answer.

Explanation:

Entropy is defined as the degree of randomness that is present within the particles of a substance.

As $NH_(4)NO_(3)$ is ionic in nature. Hence, when it is added to water then it will readily dissociate into ammonium ions ( $NH^}{+}_(4)$ ) and nitrate ions ( $NO^(-)_(3)$ ).

Therefore, it means that ions of ammonium nitrate will be free to move from one place to another. Hence, there will occur an increase in entropy.

Thus, we can conclude that ammonium nitrate ( $NH_(4)NO_(3)$ ) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol because the overall entropy of the system increases upon dissolution of this strong electrolyte.

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The following data is given to you about a reaction you are studying: Overall reaction: 2A  D Proposed mechanism: Step 1 A + B  C (slow) Step 2 C + A  D + B (fast) [A]o = 0.500 M [B]o = 0.0500 M [C]o = 0.500 M [D]o = 1.50 M This reaction was run at a series of temperatures and it was found that a plot of ln(k) vs 1/T (K) gives a straight line with a slope of -982.7 and a Y intercept of -0.0726. What is the initial rate of the reaction at 298K?

Answers

Answer : The initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

Explanation :

The Arrhenius equation is written as:

$K=A* e^{(-Ea)/(RT)}$

Taking logarithm on both the sides, we get:

$\ln k=-(Ea)/(RT)+\ln A$ ............(1)

where,

k = rate constant

Ea = activation energy

T = temperature

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor

The equation (1) is of the form of, y = mx + c i.e, the equation of a straight line.

Thus, if we plot a graph of $\ln k$ vs $(1)/(T)$ then the graph shows a straight line with negative slope. That means,

Slope of the line = $-(Ea)/(R)$

And,

Intercept = $\ln A$

As we are given that:

Slope of the line = -982.7 = $-(Ea)/(R)$

Intercept = -0.0726 = $\ln A$

Now we have to calculate the value of rate constant by putting the value of slope, intercept and temperature (298K) in equation 1, we get:

$\ln k=-(982.7)/(298)+(-0.0726)$

$\ln k=-3.37$

$k=0.0344s^(-1)$

The value of rate constant is, $0.0344s^(-1)$

Now we have to calculate the initial rate of the reaction at 298 K.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$A+B\rightarrow C$

The expression of rate law for this reaction will be,

$Rate=k[A][B]$

As we are given that:

[A] = 0.500 M

[B] = 0.0500 M

k = $0.0344s^(-1)$

Now put all the given values in the rate law expression, we get:

$Rate=(0.0344)* (0.500)* (0.0500)$

$Rate= 8.6* 10^(-4)M/s$

Therefore, the initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

What coefficients must be added to balance the following reaction?_____ Pb + _____ H3PO4 yields _____ H2 + _____ Pb3(PO4)2

A: 3, 2, 1, 1

B: 3, 2, 2, 1

C: 3, 1, 3, 1

D: 3, 2, 3, 1

Answers

D: 3, 2, 3, 1 The Pb3(PO4) comes from having 3 Pb and 2 H3PO4. The H2 comes from the 2H3PO4

Two identical containers, one red and one yellow, are inflated with different gases at the same volume and pressure. Both containers have an identically sized hole that allows the gas to leak out. It takes four times as long for the yellow container to leak out compared to the red container. If the red container is twice as hot as the yellow container, what is the ratio of the molar masses of the gases (Myellow / Mred)

Answers

Answer:

Explanation:

Here we're dealing with the root mean square velocity of gases. We'll provide the formula in order to calculate the root mean square velocity of a gas:

$v_(rms)=\sqrt{(3RT)/(M)}$

Here:

$R = 8.314 (J)/(K mol)$ is the ideal gas law constant;

$T$ is the absolute temperature in K;

$M$ is the molar mass of a compound in kg/mol.

We know that the gas from the red container is 4 times faster, as it takes 4 times as long for the yellow container to leak out, this means:

$(v_(rms, red))/(v_(rms, yellow)) = 4$

We also know that the temperature of the red container is twice as large:

$(T_(red))/(T_(yellow)) = 2$

Write the ratio of the velocities and substitute the variables:

$(v_(rms, red))/(v_(rms, yellow))=\frac{\sqrt{(3RT_(red))/(M_(red))}}{\sqrt{(3RT_(yellow))/(M_(yellow))}}=4$

Then:

$\frac{\sqrt{(3RT_(red))/(M_(red))}}{\sqrt{(3RT_(yellow))/(M_(yellow))}}=\sqrt{(3RT_(red))/(M_(red))\cdot (M_(yellow))/(3RT_(yellow))}=\sqrt{(T_(red))/(T_(yellow))\cdot (M_(yellow))/(M_(red))}=4$

From here:

$16 = (T_(red))/(T_(yellow))\cdot (M_(yellow))/(M_(red))$

Then:

$(M_(yellow))/(M_(red)) = (16)/((T_(red))/(T_(yellow))) = (16)/(2) = 8$

Final answer:

Considering Graham's Law of Effusion, and given that the temperature in the red container is twice that in yellow, the molar mass of the gas in the yellow container is 16 times that of the gas in the red container.

Explanation:

The question is about comparing the molar masses of the gases based on the rate at which they escape or effuse from two different containers. The key to this problem lies in understanding Graham's Law of Effusion, which states that the rate at which a gas effuses is inversely proportional to the square root of its molar mass.

Firstly, note that it is given that the red container takes 1/4th the time as yellow to effuse completely, meaning the gas in the red container effuses 4 times faster than the gas in the yellow container. Hence, the ratio of rates of effusion is 4:1

It is also given that the temperature in the red container is twice that in the yellow. Given the gases are in the same volume and pressure, by Graham's law, the ratio of molar masses (Myellow / Mred) would be the square of the ratio of their effusion rates, however when different temperatures are considered, it's the square of [ratio of their effusion rates x (Tred / Tyellow)].

So the ratio of the molar mass of the yellow container to the red would be (4*22)2 = 16, implying that the molar mass of the gas in the yellow container is 16 times that of the gas in the red container.

Learn more about Graham's Law here:

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Calculate your experimentally determined percent mass of water in Manganese(II) sulfate monohydrate. Report your result to 2 or 3 significant figures, e. g. 9.8% or 10.2%.

Answers

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

What is percentage mass?

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

$\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} * 100 \%$

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

$\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} * 100 \%\n\n\text{Percent Mass} = 10.6 \%}$

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

Learn more about percentage mass:

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Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of

hydration must be included.

1 Manganes 52.94 g = 63.55 g

1 Sulphur 32.07 g =

32.07 g 2 Hydrogen is = 2.02 g

4 Oygen =

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol 18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

All the elements beyond uranium, the transuranium elements, have been prepared by bombardment and are not naturally occurring elements. The first transuranium element neptunium, NpNp, was prepared by bombarding U−238U−238 with neutrons to form a neptunium atom and a beta particle. Part A Complete the following equation: 10n+23892U→?+?01n+92238U→?+? Express your answer as a nuclear equation.

Answers

Answer:

¹₀n+ ²³⁸₉₂U → ²³⁹₉₃Np + ⁰₋₁e

Explanation:

Key statement;

The first transuranium element neptunium, NpNp, was prepared by bombarding U−238U−238 with neutrons to form a neptunium atom and a beta particle.

This is the beta particle; ⁰₋₁e

¹₀n+ ²³⁸₉₂U → Np + ⁰₋₁e

The mass number of Np;

1 + 238 = Np + 0

Np = 239

The atomic number of Np;

0 + 92 = Np + (-1)

92 + 1 = Np

Np = 93

The equation is given as;

¹₀n+ ²³⁸₉₂U → ²³⁹₉₃Np + ⁰₋₁e

Be sure to answer all parts. Consider both 5-methyl-1,3-cyclopentadiene (A) and 7-methyl-1,3,5-cycloheptatriene (B). Which labeled H atom is most acidic? Hb is most acidic because its conjugate base is aromatic. Hc is most acidic because its conjugate base is antiaromatic. Ha is most acidic because its conjugate base is antiaromatic. Hd is most acidic because its conjugate base is aromatic. Which labeled H atom is least acidic? Ha is least acidic because its conjugate base is aromatic. Hb is least acidic because its conjugate base is antiaromatic. Hd is least acidic because its conjugate base is aromatic. Hc is least acidic because its conjugate base is antiaromatic.

Answers

Due to the conjugate base of the hydrogen atom is aromatic, Hb is regarded as the most acidic. Because the conjugate base of the hydrogen atom Hc is anti-aromatic, it is the least acidic.

The correct options are:

(A) - (a)

(B) - (d)

What are the most and the least acidic hydrogen atom?

The hydrogen connected at the heptatriene's tertiary position (at the 7-methyl) would be particularly acidic, as its removal would leave a positive charge that could be transported around the ring via resonance.

The hydrogen connected to the pentadiene (5-methyl) at the tertiary position would not be acidic, as removing it would result in an anti-aromatic structure.

Thus, the least acidic H atom is Hc and the most acidic H atom is Hb.

Learn more about hydrogen atom, here:

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I don’t have a picture but I can describe it to you.

The hydrogen that is attached at the tertiary position on the heptatriene (at the 7-methyl) would be very acidic, as removal would leave a positive charge that could be moved throughout the ring through resonance. This would mean that the three double bonds would be participating in resonance, and the deprotonated structure would be aromatic, thus making this favorable.

The hydrogen that is attached at the tertiary position on the pentadiene (5-methyl) would NOT be acidic, as removal would cause an antiaromatic structure.

Any other hydrogens would NOT be acidic. Those vinylic to their respective double bonds would seriously destabilize the double bond if removed, and hydrogens attached to the methyl group jutting off the ring have no incentive to leave the carbon.

Hope this helps!

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