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Alisiya [41]
2 years ago
8

What is the total mass of 2.0 moles of h2( g.??

Chemistry
1 answer:
irinina [24]2 years ago
6 0
When we speak of hydrogen, the assumption is that it is hydrogen gas, H2, and not atomic hydrogen. Since hydrogen is diatomic, 1 mol of H2 has a mass of 2.0 grams. Therefore, 2 moles will have a mass of 4.0 grams.

Likewise, 0.5 grams is 1/4 of the mass of 1 mole (2.0 g), so the answer is 0.25 mol H2.

You can even set up a conversion factor to get the answer.
0.5 g H2 x (1 mol H2 / 2.0 g H2) = 0.25 mol H2
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Find the total surface area of the rectangular prism in the figure. A) 174 cm^ 2 B) 522c * m ^ 2; 348c * m ^ 2 D) 432c * m ^ 2
WARRIOR [948]
Surface area: 348 m^2
8 0
1 year ago
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NOT A TEST AND I WILL GIVE BRAINLEIST DON'T ANSWER IF YOU DON'T KNOW
barxatty [35]

Answer:

In the nucleus of an atom ,there are protons & neutrons. Protons have charge of 1.6× 10^-19 C, while neutrons have 0C charges. Electrons orbit outside the atom. Their charge is - 1.6 ×10^-19C

Explanation:

5 0
2 years ago
Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if
mrs_skeptik [129]

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

  • <em>initial volume of gas in the container, V₁ = 2.65 L</em>
  • <em>initial number of moles of gas, n₁ = 0.12 mol</em>
  • <em>additional concentration, n = 0.182 mol</em>

The total number of moles of gas in the container is calculated as follows;

n_t = 0.12 + 0.182 = 0.302 \ mol

The final volume of gas in the container is calculated as follows;

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L

Thus, the final volume of the methane gas in the container is 6.67 L.

Learn more here:brainly.com/question/21912477

5 0
2 years ago
In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0
2 years ago
Is it called when a substance enters a gaseous phase without boiling?
Helen [10]

Answer:

its called sublimation

3 0
3 years ago
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