Balance the following equation using the smallest whole number coefficients:

Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the chemical equation shown below.

UNO [17]

Answer: The enthalpy change of the reaction is -361.6 kJ

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of sodium = 0.025 moles

Molar mass of sodium = 23 g/mol

Putting values in above equation, we get:

$0.025mol=\frac{\text{Mass of sodium}}{23g/mol}\\\\\text{Mass of sodium}=(0.025mol\times 23g/mol)=0.575g$

We are given:

Mass of water = 100.00 g

Mass of sodium = 0.575 g

Mass of solution = 100.00 + 0.575 = 100.575 g

To calculate the amount of heat absorbed, we use the equation:

$q=m\times C\times \Delta T$

where,

q = amount of heat absorbed = ?

m = mass of solution = 100.575 g

C = specific heat capacity of solution = 4.18 J/g°C

$\Delta T$ = change in temperature = $(T_2-T_1)=(35.75-25.00)=10.75^oC$

Putting all the values in above equation, we get:

$q=100.575g\times 4.18J/g^oC\times 10.75^oC=4519.34J=4.52kJ$

When heat is absorbed by the solution, this means that heat is getting released by the reaction.

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

$2Na(s)+2H_2O(l)\rightarrow NaOH(aq.)+H_2(g)$

When 0.025 moles of sodium is reacted, the heat released by the reaction is 4.52 kJ

So, when 2 moles of sodium will react, the heat released by the reaction will be = $\frac{4.52}{0.025}\times 2=361.6kJ$

Hence, the enthalpy change of the reaction is -361.6 kJ

6 0

3 years ago

How does this process appear in the atmosphere?

Mariulka [41]

D.The transferring of thermal energy from the sun to the earth!

8 0

3 years ago

Read 2 more answers

For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)

aev [14]

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

7 0

3 years ago

calculate the concentration in parts per million (ppm) of DDT if a sample size of 2000 g contained 0.050 g DDT

Firlakuza [10]

Answer:

= 25 ppm

Explanation:

PPM also refers to parts per million, it represents a low concentration of a solution. It represents 0.001 gram or a milligram in a 1000 mL, which equivalent to 1 mg per liter.

Given; a sample size of 2000 g contained 0.050 g DDT

It means, 2000 mL sample contained 50 mg DDT

Therefore in ppm we get;

= 50 mg/ 2 L

= 25 mg/L

= 25 ppm

4 0

3 years ago

A solution contains 0.021 M Cl and 0.017 M I. A solution containing copper (I) ions is added to selectively precipitate one of t

lidiya [134]

Answer: Copper (I) iodide will precipitate first.

Explanation:

We are given:

$K_{sp}$ of CuCl = $1.0\times 10^{-6}$

$K_{sp}$ of CuI = $5.1\times 10^{-12}$

Concentration of $Cl^-\text{ ion}=0.021M$

Concentration of $I^-\text{ ion}=0.017M$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

For CuCl:

$K_{sp}=[Cu^+][Cl^-]$

Putting values in above equation, we get:

$1.0\times 10^{-6}=[Cu^+]\times 0.021$

$[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M$

Concentration of copper (I) ion = $4.76\times 10^{-5}M$

For CuI:

$K_{sp}=[Cu^+][I^-]$

Putting values in above equation, we get:

$5.1\times 10^{-12}=[Cu^+]\times 0.017$

$[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M$

Concentration of copper (I) ion = $3.00\times 10^{-10}M$

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.

7 0

4 years ago