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4 years ago

Which of the following represents the mass of 1 molecule of SH2?

Chemistry

2 answers:

alexandr402 [8]4 years ago

6 0

Answer:34.1amu

Explanation:

mr_godi [17]4 years ago

5 0

Answer:
5.645 × 10⁻²³ g

Solution:

Step 1) Calculate Molar Mass of SH₂;

Atomic Mass of Sulfur = 32 g/mol

Atomic Mass of H₂ = 2 g/mol
--------------------
Molecular Mass of SH₂ = 34 g/mol

Step 2: Calculate mass of one molecule of SH₂ as;

As,

Moles = # of Molecules / 6.022 × 10²³

Also, Moles = Mass / M.Mass So,

Mass/M.mass = # of Molecules / 6.022 × 10²³

Solving for Mass,

Mass = # of Molecules × M.mass / 6.022 × 10²³

Putting values,

Mass = (1 Molecule × 34 g.mol⁻¹) ÷ 6.022 × 10²³

Mass = 5.645 × 10⁻²³ g

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A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago

1. A sample of oxygen is collected over water at 22 ° C and 762 torr. What is the partial pressure of the dry oxygen? The vapor

Rom4ik [11]

Answer: The partial pressure of the dry oxygen is 742 torr

Explanation:

Dalton's Law of Partial Pressure states that the total pressure exerted by a mixture of gases is the sum of partial pressure of each individual gas present. Thus $P(total)=P_1+P_2 .........$