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2 years ago

A cylinder of argon contains 50 L of Ar at 12.4 atm and 127°C . How many moles of argon are in the cylinder

Chemistry

1 answer:

joja [24]2 years ago

6 0

Answer:

18.9 moles

Explanation:

We have the following data:

V = 50 L

P = 12.4 atm

T= 127°C + 273 = 400 K

R = 0.082 L.atm/K.mol (it is the gas constant)

We use the ideal gas equation to calculate the number of moles n of the gas:

PV = nRT

⇒ n = PV/RT = (12.4 atm x 50 L)/(0.082 L.atm/K.mol x 400 K) = 18.9 mol

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How many moles of 0.225 M CaOH2 are present in 0.350 L of solution?

weeeeeb [17]

Answer : The number of moles of solute $Ca(OH)_2$ is, 0.0788 moles.

Explanation : Given,

Molarity = 0.225 M

Volume of solution = 0.350 L

Formula used:

$\text{Molarity}=\frac{\text{Moles of }Ca(OH)_2}{\text{Volume of solution (in L)}}$

Now put all the given values in this formula, we get:

$0.225M=\frac{\text{Moles of }Ca(OH)_2}{0.350L}$

$\text{Moles of }Ca(OH)_2=0.0788mol$

Therefore, the number of moles of solute $Ca(OH)_2$ is, 0.0788 moles.

7 0

3 years ago

Which of the following is an anion? A. O^2- C. Al^3+ D. H

kifflom [539]

The correct answer is A. O^2-

6 0

2 years ago

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

The activation energy for the decomposition = 33813.28 J/mol

8 0

3 years ago

Which of the following is not true of "concentration"?

lions [1.4K]

The correct answer is B. The concentration of a solution does not decreases when you add more solute to the solvent. Instead, the concentration increases. Concentration is expressed as the amount of solute per unit of solvent. Therefore, increasing the solute, increases this value and increasing the solvent, decreases this value.

6 0

3 years ago

20. You are approaching a downhill. Which configuration woulo

LenKa [72]

Answer: I think the answer is A

Explanation:

I’m sorry if I’m wrong.

4 0

2 years ago