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ollegr [7]
3 years ago
8

A sample of an unknown gas with a mass of 8.21 g has a volume of 4.8064 L when the temperature is 200oC and the pressure is 1.81

6 atm. What is the molar mass of the gas?
Chemistry
1 answer:
Pie3 years ago
4 0

Answer:

The answer to your question is M = 36.49 g

Explanation:

Data

mass = 8.21 g

volume = 4.8064 L

Temperature = 200°C

Pressure = 1.816 atm

M = ?

Process

1.- Convert temperature to °K

°K = 273 + 200

°K = 473

2.- Calculate the number of moles

n = (PV)/RT

n = (1.816)(4.8064)/(0.082)(473)

n = 0.225

3.- Calculate the molar mass

            M --------------- 1 mol

           8.21 g ---------- 0.225 moles

M = (1 x 8.21)/0.225

M = 36.49 g

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A 46.2 mL,0.568 M calcium nitrate solution is mixed with 80.5mL of 1.396M calcium nitrate solution.Calculate tge concentration o
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Answer:

1.09 M

Explanation:

Let's define the equation that will be used to calculate the final concentration of the resultant calcium nitrate solution. In order to calculate it, we need to find the total number of moles of calcium nitrate and divide by the total volume of the resultant solution:

c=\frac{n}{V}

This equation firstly helps us find the number of moles of calcium nitrate. Multiplying molarity by volume will yield the moles. Adding the moles from the first component to the second component will provide us with the total number of moles of calcium nitrate:

n_{Ca(NO_3)_2}=46.2 mL\cdot0.568 M+80.5 mL\cdot1.396 M=138.62 mmol

Now, the total volume of this solution can be found by adding the volume values of each component:

V_total=46.2 mL+80.5 mL=126.7 mL

Finally, dividing the moles found by the total volume will yield the final molarity:

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3 years ago
HELLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPP
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Answer: The coefficients for the given reaction species are 1, 6, 2, 3.

Explanation:

The given reaction equation is as follows.

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Now, the two half-reactions can be written as follows.

Reduction half-reaction: Cr_{2}O^{2-}_{7} + 3e^{-} \rightarrow Cr^{3+}

This will be balanced as follows.

Cr_{2}O^{2-}_{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O ... (1)

Oxidation half-reaction: Cl^{-} \rightarrow Cl_{2} + 1e^{-}

This will be balanced as follows.

6Cl^{-} \rightarrow 3Cl_{2} + 6e^{-} ... (2)

Adding both equation (1) and (2) we will get the resulting equation as follows.

Cr_{2}O^{2-}_{7} + 14H^{+} + 6Cl^{-} \rightarrow 2Cr^{3+} + 3Cl_{2} + 7H_{2}O

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