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Dafna11 [192]
3 years ago
12

When the bromine atom becomes an anion, it ______ in size.

Chemistry
1 answer:
GenaCL600 [577]3 years ago
3 0

Explanation:

Anions and cations have different atomic sizes than the neutral atoms.

When the bromine atom becomes an anion, it ______ in size.

Anions involves gaining extra electrons. This basically leads to increased repulsion between the electrons and thereby increasing the size. Anions are always larger than the neutral atoms.

This means the answer is A. Increase

When the strontium atom becomes a cation, it ______ in size.

Cations involves losing valence electrons. This basically leads to a decrease in electron shells and electron repulsion thereby reducing the size. Cations are always larger than the neutral atoms.

This means the answer is B. Decrease

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ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).

ΔrH = 180.6 kJ.

2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).

ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).

ΔS = 24.8 J/K.

3) ΔG = ΔH - TΔS.

55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.

2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.

3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.

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Give 3 examples of solutions, suspensions and colloids..
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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro
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<u>Answer:</u> The percent yield of water is 46.9 %

<u>Explanation:</u>

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

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

  • <u>For sulfuric acid:</u>

Given mass of sulfuric acid = 72.6 g    (Assuming)

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol

  • <u>For NaOH:</u>

Given mass of NaOH = 77 g      (Assuming)

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{77g}{40g/mol}=1.925mol

The chemical equation for the reaction of sulfuric acid and sodium hydroxide follows:

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

0.741 moles of sulfuric acid will react with = \frac{2}{1}\times 0.741=1.482mol of NaOH

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

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

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of water

0.741 moles of sulfuric acid will react with = \frac{2}{1}\times 0.741=1.482mol of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.67g

  • To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 12.5 g  (Assuming)

Theoretical yield of water = 26.67 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{12.5g}{26.67g}\times 100\\\\\% \text{yield of water}=46.9\%

Hence, the percent yield of water is 46.9 %

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