Answer:
The final volume is 39.5 L = 0.0395 m³
Explanation:
Step 1: Data given
Initial temperature = 200 °C = 473 K
Volume = 0.0250 m³ = 25 L
Pressure = 1.50 *10^6 Pa
The pressure reduce to 0.950 *10^6 Pa
The temperature stays constant at 200 °C
Step 2: Calculate the volume
P1*V1 = P2*V2
⇒with P1 = the initial pressure = 1.50 * 10^6 Pa
⇒with V1 = the initial volume = 25 L
⇒with P2 = the final pressure = 0.950 * 10^6 Pa
⇒with V2 = the final volume = TO BE DETERMINED
1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2
V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)
V2 = 39.5 L = 0.0395 m³
The final volume is 39.5 L = 0.0395 m³
1 atm = 760mmHg
754.3 mmHg / 760 mmHg * 1atm = 0.99 atm
760 mmHg = 101.3 KPa
754.3 mmHg/ 760mmHg *101.3 KPa = 100.54 KPa
Hope this helps!
The balanced equation for the above reaction is
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs.
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
Then pH of solution is 7
I don't think it wont be a big explosion
Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
