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Tpy6a [65]
3 years ago
12

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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From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
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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.

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3 years ago
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