Chemistry, 07.07.2019 20:30 VamPL

Atoxic gas. a, consists of 53.8% nitrogen and 46.2% carbon by mass. at 273 k and 1.01 x 10^5 pa, 1.048g of a occupies 462 cm^3. determine the empirical formula of a. calculate the moral mass of the compound and determine its molar mass

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Answer from: kcarstensen59070

The molecular formula of the compound is C2N2.

From the ideal gas equation;

PV = nRT

P = 1.01 x 10^5 Pa or 1.01 atm

V = 462 cm^3 or 0.462 L

R = 0.082 atm Lmol-1K-1

T = 273 K

n = ?

n = PV/RT

n = 1.01 atm × 0.462 L/ 0.082 atm Lmol-1K-1 × 273 K

n =0.467 /22.386

n = 0.02086 moles

Now;

Molar mass of the compound is obtained from;

number of moles = mass/molar mass

molar mass = mass/number of moles

molar mass = 1.048g/0.02086 moles

molar mass = 50.24 g/mol

The empirical formula of the compound is;

C - 53.8%/12, N - 46.2%/14

C - 4.48, N - 3.3

Dividing through by the lowest ratio

C -4.48/3.3 N - 3.3/3.3

C - 1 N - 1

The empirical formula is CN

The molecular formula is obtained thus;

[12 + 14]n = 50.24

n = 50.24/26

n = 2

The molecular formula of the compound is C2N2

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Answer from: Yungnaudie

By the mass percentage, 100 grams of the sample would contain

$53.8 \; \text{g}$ of nitrogen, and $46.2 \; \text{g}$ of carbon.

Which corresponds to

$53.8 / 14.01 \approx 3.84 \; \text{mol}$ of nitrogen atoms, and $46.2 / 12.01 \approx 3.85 \; \text{mol}$ of carbon atoms.

Hence the approximate ratio $n(\text{N}) : n(\text{C}) \approx 1 : 1$ and the empirical formula $\text{CN}$ .

Apply the idea gas formula to determine the number of moles of the gas molecules present in the $462 \; \text{cm}^{3} = 0.462 \; \text{L}$ sample:

$\begin{array}{lll}n &= & (\text{P}\cdot \text{V}) / (\text{R} \cdot \text{T})\\ & = & 1.01 \times 10^2 \times 0.462 /(8.314 \times 273) \\ &= &0.0206\; \text{mol}\end{array}$

Thus

$M = m / n = 1.048 / 0.0206 = 51.0 \; \text{g} \cdot \text{mol}^{-1}$

which is approximately twice of $26.02\; \text{g} \cdot \text{mol}^{-1}$ , formula mass for the empirical formula $\text{CN}$ . Hence the molecular formula $\text{C}_2\text{N}_2$

Answer from: Quest

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh $pcl_{3}$ +δh $cl_{2}$ ] - [δh $pcl_{5}$ ]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

Answer from: Quest

gas

explanation:

smoke is the matter of a gas

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Atoxic gas. a, consists of 53.8% nitrogen and 46.2% carbon by mass. at 273 k and 1.01 x 10^5 pa, 1.0...