Answer:
Q = 1267720 J
Explanation:
∴ QH2O = mCpΔT
∴ m H2O = 500 g
∴ Cp H2O = 4.186 J/g°C = 4.183 E-3 KJ/g°C
∴ ΔT = 120 - 50 = 70°C
⇒ QH2O = (500 g)(4.183 E-3 KJ/g°C)(70°C) = 146.51 KJ
∴ ΔHv H2O = 40.7 KJ/mol
moles H2O:
∴ mm H2O = 18.015 g/mol
⇒ moles H2O = (500 g)(mol/18.015 g) = 27.548 mol H2O
⇒ ΔHv H2O = (40.7 KJ/mol)(27.548 mol) = 1121.21 KJ
⇒ Qt = 146.51 KJ + 1121.21 KJ = 1267.72 KJ = 1267720 J
Answer:
The resulting solution is basic.
Explanation:
The reaction that takes place is:
First we <u>calculate the added moles of HNO₃ and KOH</u>:
- HNO₃ ⇒ 12.5 mL * 0.280 M = 3.5 mmol HNO₃
- KOH ⇒ 5.0 mL * 0.920 M = 4.6 mmol KOH
As <em>there are more KOH moles than HNO₃,</em> the resulting solution is basic.
Answer:
Explanation:
The integrated rate law for radioactive decay is
1. Calculate the decay constant
2. Calculate the half-life
Answer:..A.) Each chlorine atom shares a pair of electrons with the sulfur atom
