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3 years ago

I need help. ASAP! Look at the picture below. (2)

Chemistry

1 answer:

ozzi3 years ago

5 0

C. MgCl2

This structure forms a ionic bond, which are generally stronger than covalent bonds. All of the other choices are covalent.

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When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

yKpoI14uk [10]

Answer:

For a: The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b: The mass of nitric acid formed is 54.81 grams

For c: The mass of nitric acid formed is 206 grams

Explanation:

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

For a:

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

For c:

For nitrogen dioxide:

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

For water:

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

7 0

2 years ago

HELP ME ON NUMBER 8 ASAP

Varvara68 [4.7K]

Answer:

2. hope it helps. please follow me I'm new here

3 0

3 years ago

Calculate the standard entropy of vaporization of ethanol, C2H5OH, at 285.0 K, given that the molar heat capacity at constant pr

MArishka [77]

Answer: The standard entropy of vaporization of ethanol is 0.275 J/K

Explanation:

$C_2H_5OH(l)\rightleftharpoons C_2H_5OH(g)$

Using Gibbs Helmholtz equation:

$\Delta G=\Delta H-T\Delta S$

For a phase change, the reaction remains in equilibrium, thus $\Delta G=0$

$\Delta H=T\Delta S$

Given: Temperature = 285.0 K

$\Delta H=78.3J/mol$

Putting the values in the equation:

$78.3J=285.0K\times \Delta S$

$\Delta S=0.275J/K$

Thus the standard entropy of vaporization of ethanol is 0.275 J/K

4 0

3 years ago

UGRENT! Please help showing all work

agasfer [191]

Answer:

a. The limiting reactant is Ca(OH)₂

b. The theoretical yield of CaCl₂ is approximately 621.488 grams

c. The percentage yield of CaCl₂ is approximately 47.06%

Explanation:

a. The given chemical reaction is presented as follows;

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Therefore;

One mole of Ca(OH)₂ reacts with two moles of HCl to produce one mole of CaCl₂ and two moles of water H₂O

The mass of HCl in an experiment, m₁ = 229.70 g

The mass of Ca(OH)₂ in an experiment, m₂ = 207.48 g

The molar mass of HCl, MM₁ = 36.458 g/mol

The molar mass of Ca(OH)₂, MM₂ =74.093 g/mol

The number of moles of HCl, present, n₁ = m₁/MM₁

∴ n₁ = m₁/MM₁ = 229.70 g/(36.458 g/mol) ≈ 6.3 moles

The number of moles of Ca(OH)₂, present, n₂ = m₂/MM₂

∴ n₂ = m₂/MM₂ = 207.48 g/(74.093 g/mol) ≈ 2.8 moles

The number of moles of Ca(OH)₂, present, n₂ = 2.8 moles

According the chemical reaction equation the number of moles of HCl the 2.8 moles of Ca(OH)₂ will react with, = 2.8 × 2 moles = 5.6 moles of HCl

Therefore, there is an excess HCl in the reaction and Ca(OH)₂ is the limiting reactant

b. According the chemical reaction equation the number of moles of CaCl₂ produced in he reaction by the 2.8 moles of Ca(OH)₂ = 2.8 × 2 moles = 5.6 moles of CaCl₂

The molar mass of CaCl₂ = 110.98 g/mol

The mass of the 5.6 moles of CaCl₂ = 5.6 moles × 110.98 g/mol ≈ 621.488 grams

The theoretical yield of CaCl₂ ≈ 621.488 grams

c. Given that the actual mass of CaCl₂ produced = 292.5 grams, we have;

The percentage yield of CaCl₂ = The actual yield/(The theoretical yield) × 100

∴ The percentage yield of CaCl₂ = (292.5 g)/(621.488 g) × 100 ≈ 47.0644646397%

The percentage yield of CaCl₂ ≈ 47.06%.

8 0

3 years ago

Convert 7.50 mol of Li₂O to grams

Natali5045456 [20]

Answer:

223.5 g

Explanation:

The formula between the number of moles, mass and Mr can be used to convert moles to grams.

Number of moles = mass ÷ Mr

So, mass = number of moles × Mr

Mr of Li₂O = (6.9 × 2) + 16 = 29.8

∴ Mass = 7.5 × 29.8 = 223.5 g

4 0

2 years ago