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3 years ago

What is the molecular formula of a compound that is composed of 94.1% O and 5.9% H with a molar mass of 34g

Chemistry

1 answer:

elena-s [515]3 years ago

3 0

Answer:

B. 67.6/Hg(200.5)= .337. 10.8/S(32.1)= .336. 21.6/O(16)= 1.35--> .337/.336= 1 .336/.336= 1 1.35/.336= 4. Formula= HgSO4

Explanation:

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We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh

k0ka [10]

Answer:

The ideal molar volume is $\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The Z factor is $Z = 0.09997$

The real molar volume is $\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

Explanation:

From the question we are told that

The pressure is $P = 23 \ bar = 23 *10^5 Pa$

The temperature is $T = 30 ^ oC = 303 \ K$

According to the ideal gas equation we have that

$PV = nRT$

=> $\frac{V}{n}=V_z= \frac{RT}{P}$

Where $\frac{V}{n }$ is the molar volume and R is the gas constant with value

$R = 8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}$

substituting values

$\frac{V}{n} =V_z= \frac{ 8.314 * 303}{23 *10^{5}}$

$\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The compressibility factor of the gas is mathematically represented as

$Z = \frac{P * V_z}{RT}$

substituting values

$Z = \frac{23 *10^{5} * 0.001095}{8.314 * 303}$

$Z = 0.09997$

Now the real molar volume is evaluated as

$\frac{V_r}{n} = V_k= \frac{Z * RT }{P}$

substituting values

$\frac{V_r}{n} = V_k= \frac{0.09997 * 8.314 * 303}{23 *10^{5}}$

$\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

8 0

3 years ago

The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N2O5, dissolved in chloroform, CHCl3, is 6

weeeeeb [17]

Answer:

Rate of the reaction= 9.92× 10^-5 M² min-1

Explanation:

Using the equation of reaction

2N2O5 ⟶ 4NO2+O2

Rate = k[N2O5]²

From the question k= 6.2×10-4

[N2O5]= 0.4

Rate = 6.2×10-4[0.4]²= 9.92×10-5M² min-1

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