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faltersainse [42]

4 years ago

12

Select the correct answer from each drop-down menu. How are real gases different from ideal gases? Real gases differ from ideal

gases because in a real gas, and .

2 answers:

lawyer [7]4 years ago

8 0

They have a mass for the particles

There are no totally elastic collisions

There are intermolecular forces

ololo11 [35]4 years ago

6 0

1. Particles take up some volume

2. Intermolecular forces between particles

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1. The behavior that occurs when a wave bends as it changes media is known as

Pepsi [2]

D because its changing

4 0

3 years ago

Read 2 more answers

What is the molar mass of (NH), CO?<br> 138g<br> 788<br> 968<br> 1448

aleksandr82 [10.1K]

Answer:

The molar mass of $(NH_{4})_{2}CO_{3}$ is 96.8 g/mol

Explanation:

The given molecular formula - $(NH_{4})_{2}CO_{3}$

Individual molar masses of each element in the compound is as follows.

Molar mass of nitrogen - 14.01 g/mol

Molar mass of of hydrogen = 1.008g/mol

Molar mass of carbon = 12.01 g/mol

Molar mass of oxygen =16.00 g/mol

Molar mass of $(NH_{4})_{2}CO_{3}$ is

$2\times[1(14.01)+4(1.008)]+1(12.01)+3(16.00)= 96.8g/mol$

Therefore,The molar mass of $(NH_{4})_{2}CO_{3}$ is 96.8 g/mol

7 0

3 years ago

CaC12 * 3H20 is correctly named

ki77a [65]

Answer:

calcium chloride deihydrate

4 0

3 years ago

If 38.6 grams of iron react with an excess of bromine gas, what mass of FeBr2 can form?

Yuki888 [10]

Answer:

›› FeBr2 molecular weight. Molar mass of FeBr2 = 215.653 g/mol. This compound is also known as Iron(II) Bromide. Convert grams FeBr2 to moles or moles FeBr2 to grams. Molecular weight calculation: 55.845 + 79.904*2 ›› Percent composition by element

Explanation:

5 0

3 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For <em>hydrogen </em>we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for <em>nitrogen</em>:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

3 years ago