Calculate the number of Na ions and Cl ions and total number of ions in 14.5g of NaCl.
1 answer:
Answer:
Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.
Explanation:
1 mole of any compound contains 6.023 × 10²³ molecules.
molecular weight of NaCl is 23 + 35.5 = 58.5 g.
so, 58.5 grams of NaCl makes 1 mole
⇒ 14.5 g of NaCl = = 0.248 moles.
⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules
= 1.49 × 10²³ molecules.
And 1 molecule contains 1 Na ion and 1 Cl ion.
⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.
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Answer:
8.1107 g
Explanation:
The given reaction:
Given that:
Mass of silver sulfadiazine = 25.0 g
Molar mass of silver sulfadiazine = 357.14 g/mol
The formula for the calculation of moles is shown below:
Thus,
From the reaction,
2 moles of silver sulfadiazine are formed from 1 mole of silver oxide
So,
1 mole of silver sulfadiazine are formed from 1/2 mole of silver oxide
0.07 mole of silver sulfadiazine are formed from 1/2*0.07 mole of silver oxide
Moles of silver oxide = 0.035 moles
Molar mass of silver oxide = 231.735 g/mol
Mass = Moles * Molar mass = 0.035 moles * 231.735 g/mol = 8.1107 g