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2 years ago

PLZ HELP, GIVING BRANLIEST!!

Chemistry

1 answer:

valentinak56 [21]2 years ago

5 0

Answer:

Option A. The polar solvent molecule surrounds the positive sodium ions and the negative chloride ions.

Explanation:

When a salt say NaCl dissolved in water, the solvent molecules surround both the Na+ and Cl-. The Na+ are surrounded by OH- and the Cl- are surrounded by H+.

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Read 2 more answers

For each of the acid–base reactions, calculate the mass (in grams) of each acid necessary to completely react with and neutraliz

LiRa [457]

Acid-base neutralization reaction is defined as reaction of acid with base to form corresponding salt and water. Strong acid and base completely neutralize each other.

(a) The acid base neutralization reaction is as follows:

$HCl(aq)+NaOH(aq)\rightarrow H_{2}O(l)+NaCl(g)$

From the above balanced chemical reaction, 1 mol of NaOH completely reacts with 1 mol of HCl. The mass of NaOH is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of NaOH is 39.997 g/mol, thus,

$n=\frac{4.85 g}{39.997 g/mol}=0.121 mol$

Thus, 0.121 mol of NaOH reacts with same amount of HCl and number of moles of HCl will be 0.121 mol.

Since. molar mass of HCl is 36.46 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.121 mol\times 36.46 g/mol=4.421 g$

Therefore, 4.421 g of HCl completely reacts with 4.85 g of NaCl.

(b) The acid base neutralization reaction is as follows:

$2HNO_{3}(aq)+Ca(OH)_{2}(aq)\rightarrow 2H_{2}O(l)+Ca(NO_{3})_{2}(aq)$

From the above balanced chemical reaction, 1 mol of $Ca(OH)_{2}$ completely reacts with 2 mol of $HNO_{3}$ . The mass of $Ca(OH)_{2}$ is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of $Ca(OH)_{2}$ is 74.093 g/mol, thus,

$n=\frac{4.85 g}{74.093 g/mol}=0.06545 mol$

Thus, 0.06545 mol of $Ca(OH)_{2}$ reacts with $2\times 0.06545 mol=0.13091 mol$ of $HNO_{3}$

Since. molar mass of $HNO_{3}$ is 63.01 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.13091 mol\times 63.01 g/mol=8.25 g$

Therefore, 8.25 g of $HNO_{3}$ completely reacts with 4.85 g of $Ca(OH)_{2}$ .

$H_{2}SO_{4}(aq)+2 KOH (aq)\rightarrow 2H_{2}O(l)+K_{2}SO_{4}(aq)$

From the above balanced chemical reaction, 2 mol of KOH completely reacts with 1 mol of $H_{2}SO_{4}$ . The mass of KOH is given 4.85 g, convert this into number of moles as follows: