How many hydrogen (H) atoms are in 1 molecule of (NH4)2S ?

What is the volume of 2 mol of chlorine gas at STP? 2.0 L 11.2 L 22.4 L 44.8 L

nirvana33 [79]

Answer:

44.8 L

Explanation:

Using the ideal gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

At Standard temperature and pressure (STP);

P = 1 atm

T = 273K

Hence, when n = 2moles, the volume of the gas is:

Using PV = nRT

1 × V = 2 × 0.0821 × 273

V = 44.83

V = 44.8 L

7 0

2 years ago

Convert 7.02 mol of NaCl to formula units??

nika2105 [10]

To convert the number of moles a compound to formula units, the conversion factor is Avogadro's number, 6.022 x10^23 formula units/ mol. In this case, we are given with 7.02 moles of sodium chloride. Hence the answer is 4.23 x 10^24 formula units.

8 0

3 years ago

Problem page what kind of intermolecular forces act between a dichlorine monoxide molecule and a hydrogen bromide molecule?

marta [7]

Dichlorine monoxide has the same structure like that of water. So, this is a polar molecule. For polar molecules, the dominant intermolecular force would be dipole-dipole forces. For HBr, there is a force between two oppositely charged ions, H⁺ and Br⁻. So, the dominant intermolecular force is electrostatic attraction.

5 0

3 years ago

Read 2 more answers

Based on Charles's law, which of the following statements is true for an ideal gas at a constant pressure and mole amount?

grandymaker [24]

Answer:

Volume is directly proportional to absolute temperature.

Explanation:

5 0

3 years ago

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

3 0

3 years ago