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Lynna [10]
3 years ago
7

A gas with an empirical formula C2H2O has a molecular weight of 120.6g/mol. A possible molecular formula for the gas is

Chemistry
1 answer:
Virty [35]3 years ago
7 0

M(C2H2O)= 12.0*2 +1.0*2 +16.0 = 42 g/mol is a molar mass for empirical formula.

120.6g/mol/42g/mol ≈ 3

So, empirical formula should be increased 3 times,

and molecular formula is C6H6O3.

Answer is D.

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Read 2 more answers
8. FOR BRAINLIST!!!
Ne4ueva [31]

I think it's either b or c

5 0
3 years ago
A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som
irinina [24]

Answer:

m_{KBr}=6.030gKBr

Explanation:

Hello.

In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

(T_b-T_b_0)=i*m*K_b

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

m=\frac{T_b-T_b_0}{i*K_b}=\frac{(119.3-117.8)\°C}{2*1.48\°C*kg*mol^{-1}}\\  \\m=0.507mol/kg

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

n_{KBr}=0.507mol/kg*0.1kg=0.0507mol

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

m_{KBr}=0.0507mol*\frac{119g}{1mol} \\\\m_{KBr}=6.030gKBr

Best regards.

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4 years ago
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