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Kisachek [45]
2 years ago
15

Calculate the volume (in liters) of 12.5 g of argon gas at a pressure of 1.05 atm and a temperature of 322k.

Chemistry
1 answer:
Bess [88]2 years ago
4 0

The volume (in liters) of 12.5 g of argon gas at a pressure of 1.05 atm and a temperature of 322 k is 7.86L.

Given,

Mass of argon = 12.5 g

Molar mass of argon = 40 g

Pressure = 1.05 atm

Temperature = 322K

By using ideal gas equation:

PV = nRT

where,

P = pressure, V = volume, n = number of moles,

R = gas constant = 0.0821Latm/ mol K

and T = temperature

<h3>Calculation of moles</h3>

Moles = given mass/ molar mass

= 12.5/40

= 0.3125 mol

<h3>Calculation of volume</h3>

V = nRT/P

V = 0.3125× 0.0821 × 322/1.05

= 7.86L

Thus, we find that the volume of 12.5 g of argon gas is 7.86L.

learn more about ideal gas equation:

brainly.com/question/4147359

#SPJ4

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PLEASE HELP ASAP
swat32

Answer:

12.34 amu

Explanation:

Let the 1st isotope be A

Let the 2nd isotope be B

Let the 3rd isotope be C

From the question given above, the following data were obtained:

1st Isotope (A):

Mass of A = 12.32 amu

Abundance (A%) = 19.5%

2nd isotope (B):

Mass of B = 13.08 amu

Abundance (B%) = 26.23%

3rd isotope (C):

Mass of C = 11.99 amu

Abundance (C%) = 54.27%

Atomic mass of X =?

The atomic mass of the element X can be obtained as follow:

Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]

= [(12.32 × 19.5)/100] + [(13.08 × 26.23)/100] + [(11.99 × 54.27)/100]

= 2.402 + 3.431 + 6.507

= 12.34 amu

Thus, the atomic mass of the element X is 12.34 amu

7 0
3 years ago
You are given a stock solution of 500.0 mL of 1.00M magnesium chloride solution. Calculate the volume of the stock solution you
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Answer:

50\; \rm mL of the stock solution would be required.

Explanation:

Assume that a solution of volume V contains a solute with a concentration of c. The quantity n of that solute in this solution would be:

n = c \cdot V.

For the solution that needs to be prepared, c = 0.20\; \rm M = 0.20\; \rm mol \cdot L^{-1}. The volume of this solution is V = 250.0\; \rm mL. Calculate the quantity of the solute (magnesium chloride) in the required solution:

\begin{aligned}n &= c \cdot V \\ &= 0.20\; \rm mol \cdot L^{-1} \times 250.0\; \rm mL \\ &= 0.20\; \rm mol \cdot L^{-1} \times 0.2500\; \rm L \\ &= 0.050\; \rm mol\end{aligned}.

Rearrange the equation n = c \cdot V to find an expression of volume V, given the concentration c and quantity n of the solute:

\displaystyle V= \frac{n}{c}.

Concentration of the solute in the stock solution: c(\text{stock}) = 1.00\; \rm M = 1.00\; \rm mol \cdot L^{-1}.

Quantity of the solute required: n = 0.050\; \rm mol.

Calculate the volume of the stock solution that would contain the required n = 0.050\; \rm mol of the magnesium chloride solute:

\begin{aligned}& V(\text{stock}) \\ &= \frac{n}{c(\text{stock})} \\ &= \frac{0.050\; \rm mol}{1.00\; \rm mol \cdot L^{-1}} \\ &= 0.050\; \rm L \\ &= 50\; \rm mL\end{aligned}.

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