Part a a solution is made by mixing 9.00 mmol (millimoles) of ha and 3.00 mmol of the strong base. what is the resulting ph? exp
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Answer:
806.3g
Explanation:
Given parameters:
Number of moles of silver nitrate = 4.85mol
Unknown:
Mass of silver chromate = ?
Solution:
2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃
To solve this problem, we work from the known to the unknown;
- The known specie here is AgNO₃ ;
From the balanced chemical equation;
2 moles of AgNO₃ will produce 1 mole of Ag₂CrO₄
4.85 moles of AgNO₃ will produce = 2.43moles of Ag₂CrO₄
- Mass of silver chromate produced;
mass = number of moles x molar mass
Molar mass of Ag₂CrO₄
Atomic mass of Ag = 107.9g/mol
Cr = 52g/mol
O = 16g/mol
Input the parameters and solve;
Molar mass = 2(107.9) + 52 + 4(16) = 331.8g/mol
So,
Mass of Ag₂CrO₄ = 2.43 x 331.8 = 806.3g
