
Рассчитайте сумарное число ионов: 1) в 11,7 г натрий хлорида; 2) в 20.4 г алюминий оксида 3) в 41 г
кальция нитрата

Ответы на вопрос

1)N=(11,7/58,5)*6,02*10^23*2=4*10^23 ионов
2)в растворе не будет ионов
3)N=(41/164)*3*6,02*10^23=4,5*10^23 ионов



Calculating the Total Number of Ions
1) In 11.7 g of sodium chloride: - The molar mass of sodium chloride (NaCl) is approximately 58.44 g/mol. - To calculate the number of moles, we divide the given mass by the molar mass: - Number of moles = 11.7 g / 58.44 g/mol - The formula unit of sodium chloride consists of one sodium ion (Na⁺) and one chloride ion (Cl⁻). - Therefore, the total number of ions is twice the number of moles.
Calculation: - Number of moles = 11.7 g / 58.44 g/mol = 0.2 moles - Total number of ions = 0.2 moles * 2 = 0.4 moles
2) In 20.4 g of aluminum oxide: - The molar mass of aluminum oxide (Al₂O₃) is approximately 101.96 g/mol. - To calculate the number of moles, we divide the given mass by the molar mass: - Number of moles = 20.4 g / 101.96 g/mol - The formula unit of aluminum oxide consists of two aluminum ions (Al³⁺) and three oxide ions (O²⁻). - Therefore, the total number of ions is five times the number of moles.
Calculation: - Number of moles = 20.4 g / 101.96 g/mol = 0.2 moles - Total number of ions = 0.2 moles * 5 = 1 mole
3) In 41 g of calcium nitrate: - The molar mass of calcium nitrate [Ca(NO₃)₂] is approximately 164.09 g/mol. - To calculate the number of moles, we divide the given mass by the molar mass: - Number of moles = 41 g / 164.09 g/mol - The formula unit of calcium nitrate consists of one calcium ion (Ca²⁺) and two nitrate ions (NO₃⁻). - Therefore, the total number of ions is three times the number of moles.
Calculation: - Number of moles = 41 g / 164.09 g/mol = 0.25 moles - Total number of ions = 0.25 moles * 3 = 0.75 moles
Summary:
1) In 11.7 g of sodium chloride, the total number of ions is 0.4 moles. 2) In 20.4 g of aluminum oxide, the total number of ions is 1 mole. 3) In 41 g of calcium nitrate, the total number of ions is 0.75 moles.

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