Answer:
15) (16+2)= 18g 144g÷18 = 6moles
16) He = 2 =2moles
Answer: 223.2 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
According to the given balanced equation:
4 moles of Aluminium reacts with 3 moles of oxygen .
Thus 9.30 moles of Aluminium reacts with= of oxygen.
Mass of Oxygen gas = Moles × Molar mass = 6.975mole × 32 g/mol = 223.2 g
Answer :
The atmospheric pressure of water if boiling at 81 °C will be 369 mmHg.
Explanation : The vapor pressure of boiling water can be calculated using the Antoine Equation:
log P = A - ,
where P is the pressure in mmHg,
A = 8.07131,
B = 1730.63
C = 233.426
and T = temperature in °C (81°C)
Substituting the given values, we get,
Log P = 8.07131−
Therefore, Log P =8.07131 − 5.5041 = 2.567.
Now, P = =369 mmHg.
Hence, the atmospheric pressure will be 369 mmHg.
Answer: 8.45 L
Explanation:
Given that,
Initial volume (V1) = 3.5L
Initial pressure (P1) = 2.5 atm
[Since final pressure is given in torr, convert 2.5 atm to torr
If 1 atm = 760 torr
2.5 atm = 2.5 x 760 = 1900 torr
Final volume (V2) = ?
Final pressure (P2) = 787 torr
Since pressure and volume are given while temperature remains the same, apply the formula for Boyle's law
P1V1 = P2V2
3.5L x 1900 torr = 787 torr x V2
6650L•torr = 787 torr•V2
Divide both sides by 787 torr
6650L•torr/787 torr = 787 torr•V2/787 torr
8.45 L = V2
Thus, the volume of the gas at 787 torr and at the same temperature is 8.45 Liters
Answer:
209.3 Joules require to raise the temperature from 10 °C to 15 °C.
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m × c × ΔT
Given data:
mass of water = 10 g
initial temperature T1= 10 °C
final temperature T2= 15 °C
temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C
Energy or joules added to increase the temperature Q = ?
Solution:
We know that specific heat of water is 4.186 J/g .°C
Q = m × c × ΔT
Q = 10 g × 4.186 J/g .°C × 5 °C
Q = 209.3 J