<h2><u>
Answer:</u></h2>
(These are not rounded to the correct decimal)
130.94 atm
13,266.6 kPa
99,571.4 mmHg
<h2><u>
Explanation:</u></h2>
<u></u>
PV = nRT
V = 245L
P = ?
R = 0.08206 (atm) , 8.314 (kPa) , 62.4 (mmHg)
T = 273.15 + 27 = 300.15K
n = 1302.5 moles
How I found (n).
5.21kg x 1000g/1kg x 1 mole/4.0g = 1302.5 moles
Now, plug all the numbers into the equation.
Pressure in atm = (1302.5)(0.08206)(300.15) / 245 = 130.94 atm (not rounded to the correct decimal)
Pressure in kPa = (1302.5)(8.314)(300.15) / 245 = 13,266.6 kPa (not rounded to the correct decimal)
Pressure in mmHg = (1302.5)(62.4)(300.15) / 245 = 99,571.4 mmHg (not rounded to the correct decimal)
48.3 g AgNO3 / 169.9 g/mol = 0.284 moles AgNO3
0.284 mol AgNO3 X (1 mol Ag2CrO4/2 mol AgNO3) = 0.142 mol Ag2CrO4
0.142 mol Ag2CrO4 X 331.7 g/mol = 47.1 g Ag2CrO4
1 mols of Aluminium ion forms 1 mol aluminium phosphate
Molar mass of AlPO_4
Moles of AlPO_4
- 61µg/106
- 0.000061/106
- 5.75×10^{-7}
- 57.5µmol
Moles of Al3+=57.5µmol
Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
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