a. pH=2.07
b. pH=3
c. pH=8
<h3>Further explanation</h3>
pH=-log [H⁺]
a) 0.1 M HF Ka = 7.2 x 10⁻⁴
HF= weak acid
![\tt [H^+]=\sqrt{Ka.M}\\\\(H^+]=\sqrt{7.2.10^{-4}\times 0.1}\\\\(H^+]=8.5\times 10^{-3}\\\\pH=3-log~8.5=2.07](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D%5Csqrt%7B7.2.10%5E%7B-4%7D%5Ctimes%200.1%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D8.5%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~8.5%3D2.07)
b) 1 x 10⁻³ M HNO₃
HNO₃ = strong acid
![\tt pH=-log[1\times 10^{-3}]=3](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-3%7D%5D%3D3)
c) 1 x 10⁻⁸ M HCl
![\tt pH=-log[1\times 10^{-8}]=8](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-8%7D%5D%3D8)
Answer:
Mass = 11.78 g of P₄
Explanation:
The balance chemical equation is as follow:
6 Sr + P4 → 2 Sr₃P₂
Step 1: Calculate moles of Sr as;
Moles = Mass / M/Mass
Moles = 50.0 g / 87.62 g/mol
Moles = 0.570 moles
Step 2: Find moles of P₄ as;
According to equation,
6 moles of Sr reacted with = 1 mole of P₄
So,
0.570 moles of Sr will react with = X moles of P₄
Solving for X,
X = 1 mol × 0.570 mol / 6 mol
X = 0.0952 mol of P₄
Step 3: Calculate mass of P₄ as,
Mass = Moles × M.Mass
Mass = 0.0952 mol × 123.89 g/mol
Mass = 11.78 g of P₄
Explanation:
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Answer:
It might be responding variable.