Given:
n = 12 moles of oxygen
T = 273 K, temperature
p = 75 kPa, pressure
Use the ideal gas law, given by
where
V = volume
R = 8.3145 J/(mol-K), the gas constant
Therefore,
Answer: 0.363 m³
Answer:
Heat transfer = Q = 62341.6 J
Explanation:
Given data:
Heat transfer = ?
Mass of water = 50.0 g
Initial temperature = 30.0°C
Final temperature = 55.0°C
Specific heat capacity of water = 4.184 J/g.K
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 55.0°C - 30.0°C
ΔT = 25°C (25+273= 298 K)
Q = 50.0 g × 4.184 J/g.K ×298 K
Q = 62341.6 J
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol
Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol
Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???
given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.
85.2 are in a 2 L solution that has a concentration of 0.6 mol/L
Answer:
divide the volume value by 1000
So 3828/1000=3.828
