Use the Ideal Gas Law to find the moles of gas first.
Be sure to convert T from Celsius to Kelvin by adding 273.
Also I prefer to deal with pressure in atm rather than mmHg, so divide the pressure by 760 to get it in atm.
PV = nRT —> n = PV/RT
P = 547 mmHg = 547/760 atm = 0.720 atm
V = 1.90 L
T = 33°C = 33 + 273 K = 306 K
R = 0.08206 L atm / mol K
n = (0.720 atm)(1.90 L) / (0.08206 L atm / mol K)(306 K) = 0.0545 mol of gas
Now divide grams by mol to get the molecular weight.
3.42 g / 0.0545 mol = 62.8 g/mol
The answer is gonna be the last one :)
ºC = K - 273
ºC = 96 - 273
= -177ºC
answer A
hope this helps!
An increase in the number of gas molecules in the same volume container increases pressure. A decrease in container volume increases gas pressure. An increase in temperature of a gas in a rigid container increases the pressure.
The image of the bonds are missing, so i have attached it.
Answer:
A) - Sigma bond
-Sp³ and Sp³
- None
B) - Sigma and pi bond
- Sp² of C and p of O
- p of C and P of O
Explanation:
A) For compound 1;
- the molecular orbital type is sigma bond due to the end-to-end overlapping.
- Atomic orbitals in the sigma bond will be Sp³ and Sp³
- Atomic orbitals in the pi bond would be nil because there is no pi bond.
B) For compound 2;
- the molecular orbital type is sigma and pi bond
-Atomic orbitals in the sigma bond would be Sp² of C and p of O
- The Atomic orbitals in the pi bond will be; p of C and p of O
