Answer:
2.23 x 10^-4 mol
Explanation:
0°C =273 K
PV= nRT
n= PV/RT
n= (5.00 atm * 0.0010 L)/ (0.0821 * 273 K)
n= 2.23 x 10^-4
Here, we are required to determine the pressure of the sample of gas trapped in the open-tube mercury manometer shown below if Atmospheric pressure is 767 mmHg and pressure head, h = 8.9 cmHg.
- The pressure of the sample of gas trapped in the open-tube mercury manometer is;. 856 mmHg
According to the question;
- The atmospheric pressure is 767 mmHg
- The gauge pressure is 8.9 cmHg = 89mmHg.
The absolute pressure, P(abs);
is given mathematically as;
Absolute pressure = Atmospheric pressure + gauge pressure.
P(abs) = P(atm) + P(gauge)
P(abs) = 767mmHg + 89mmHg.
P(abs) = 856 mmHg.
Read more:
brainly.com/question/17200230
Answer:
NaOH = 40 g/mole
M = 10*d*m% /MW
M = 10*1.116*15/ 40
M = 4.185
––––––––––––––––––––
pH = 11 –––> pOH = 3 –––> [OH–] = 10^–3 M
M1*V1 = M2*V2
4.185* V1 = 10^–3 * 5.3
V1 = 1.27×10^–3 L = 1.27 ml
The answer would be the aftershock