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Andreyy89
2 years ago
9

Determine the mass of KNO3 that dissolves in 100. grams of water at 40 degrees celsius to produce a saturated solution

Chemistry
1 answer:
77julia77 [94]2 years ago
4 0
This is the reference data.
100 g H₂O, 40°C
c=63,9 g/100 g w.
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abolitionist movement

Explanation:

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Whats principle of Aristotle's do scientist use today
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His observation and census.
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For which process is ΔS negative? Group of answer choices A. grinding a large crystal of KCl to powder B. raising the temperatur
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Answer:

C.

Explanation:

The entropy (S) is the measure of the randomness of a system, and ΔS = Sfinal - Sinitial. As higher is the disorder of the system, as higher is the entropy.

A. When KCl is fractionated in power, there'll be more portions of it, so, the disorder must be higher, then ΔS is positive.

B. As higher is the temperature, higher is the kinetic energy of the system, and because of that, the disorder is also higher, so ΔS is positive.

C. The decrease in the volume (compression) decreases the distance between the molecules, so the system will be more organized, then ΔS is negative.

D. The volume before the mixing will be higher, and the ethanol will dissociate, so it will be more particles, the disorder will increase, and ΔS is positive.

E. Sgas > Sliquid > Ssolid because of the disorder of the molecules, then ΔS is positive.

5 0
3 years ago
Given 6193 mL of a gas at 62.3 °C. What is its volume at 38.1 °C?
Luden [163]

Answer:

5746.0 mL.

Explanation:

We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

<em>V₁T₂ = V₂T₁</em>

<em></em>

V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.

V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.

<em>∴ V₂ = V₁T₂/T₁ </em>= (6193.0 mL)(311.1 K)/(335.3 K) = <em>5746.0 mL.</em>

5 0
3 years ago
The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react
murzikaleks [220]

<u>Answer:</u> The value of K_{eq} is 4.84\times 10^{-5}

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of ammonia = \frac{0.0280}{1.00}=0.0280M

Concentration of oxygen gas = \frac{0.0120}{1.00}=0.0120M

The given chemical equation follows:

                  4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)

<u>Initial:</u>        0.0280        0.0120

<u>At eqllm:</u>    0.0280-4x   0.0120-3x   2x       6x

We are given:

Equilibrium concentration of nitrogen gas = 3.00\times 10^{-3}M=0.003

Evaluating the value of 'x', we get:

\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M

Now, equilibrium concentration of ammonia = 0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M

Equilibrium concentration of oxygen gas = 0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M

Equilibrium concentration of water = 6x=(6\times 0.0015)]=0.009M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}

Putting values in above expression, we get:

K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}

Hence, the value of K_{eq} is 4.84\times 10^{-5}

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3 years ago
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