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tatuchka [14]
3 years ago
5

A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)

Chemistry
1 answer:
elixir [45]3 years ago
8 0

Answer: \Delta H^{0}=-173.72 kJ/mol

Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.

The standard enthalpy of formation of HCl is calculated as:

\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}

Ca(OH)_{2}_{(aq)}+2HCl_{(aq)} → CaCl_{2}_{(s)}+2H_{2}O_{(l)}

Standard Enthalpy of formation for the other compounds are:

Calcium Hydroxide: \Delta H^{0}= -1002.82 kJ/mol

Calcium chloride: \Delta H^{0}= -795.8 kJ/mol

Water: \Delta H^{0}= -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]

-17.2=-1367.46+1002.82-2\Delta H

2\Delta H=17.2-364.64

\Delta H=-173.72

So, the standard enthalpy of formation of HCl is -173.72 kJ/mol

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The thin outer layer of Earth, called the crust, contains only 0.50 percent of Earth's total mass and yet is the source of almos
Rzqust [24]

Answer:

The mass of silicon in kilograms in Earth's crust is7.2729\times 10^{21} kg.

Explanation:

Mass of Earth =5.9\times 10^{21} tons

(1 ton= 2000 lb)

(1 lb =453.6 g)

1 ton = 2000 × 453.6 g =907,200 g

Mass of Earth =5.9\times 10^{21}\times 907,200 g=5.3524\times 10^{27} g

Percentage of earth crust = 0.50%

Mass of earth crust = M

0.50\%=\frac{M}{5.3524\times 10^{27}}\times 100

M=2.6762\times 10^{25} g

Percentage of the silicon in Earth's crust = 27.2 %

Mass of silicon in in Earth's crust = m

Si\%=\frac{m}{M}\times 100

27.2\%=\frac{m}{2.6762\times 10^{25} g}\times 100

m = 7.2792\times 10^{24} g=7.2729\times 10^{21} kg

1000 g = 1 kg

The mass of silicon in kilograms in Earth's crust is7.2729\times 10^{21} kg.

7 0
3 years ago
What intermolecular force attracts two non polar molecules to each other
Salsk061 [2.6K]

The intermolecular force that attracts two nonpolar molecules is London dispersion forces, which are also called induced dipole-induced

3 0
3 years ago
Identify the catalyst in this reaction, explain how
telo118 [61]

Question:

Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reaction of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps:

equations

Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.

Answer:

NO

It is present but not consumed

NO Lowers the activation energy of the reaction

Explanation:

A catalyst is a substance that is present in a chemical reaction and enables the reaction to occur at a faster rte but does not take part n the reaction

Therefore, whereby NO is not consumed, it is the catalyst

It functions by lowering the activation energy

5 0
3 years ago
A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo
Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g   (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = 62.364\text{ L.mmHg }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0
3 years ago
14.A sample of fluorine gas has a density of _____
yarga [219]

Answer:

d = 0.793 g/L

Explanation:

Given data:

Density of fluorine gas = ?

Pressure of gas = 0.554 atm

Temperature of gas = 50 °C (50+273.15K = 323.15 K)

Solution:

Formula:

PM = dRT

M = molar mass of gas

P = pressure

R = general gas constant

T = temperature

d = PM/RT

d = 0.554 atm × 37.99 g/mol / 0.0821 atm.L /mol.K × 323.15 K

d = 21.05 atm.g/mol/26.53 atm.L /mol

d = 0.793 g/L

8 0
3 years ago
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