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3 years ago

What is the molar mass of an unknown hydrocarbon whose density is measured to be 1.97 g/l at stp?

Chemistry

2 answers:

notka56 [123]3 years ago

8 0

Answer:

The correct answer is

M = 44.1 g / mol

Explanation:

Hello!

Let's solve this!

First we write down the data

d: 1.97 g / L

P: 1 atm

R: 0.082 atmL / Kmol

T: 273K

From the ideal gas formula:

d = P * M / R * T

We cleared M to know the molar mass.

M = (d * R * T) / P

M = (1.97 g / L * 0.082 atmL / Kmol * 273K) / 1atm

The correct answer is

M = 44.1 g / mol

inessss [21]3 years ago

5 0

Answer: The correct answer is : 44.1 g / mol

Explanation: d = PM / RT

M = dRT / P

d = 1.97 g / l

R = ideal gas (0.082 06 l - atm / mol - ° K)

P = 1 atm

T = 273 ° K

M = Molar mass

M = 1.97 g / l x 0.082 l - atm / mol - ° K x 273 ° K / 1 atm

M = 44.1 g / mol

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Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago