Answer:
10 atm.
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question,
P1 = 5 atm
P2 = ?
V1 = 4L
V2 = 2L
T1 = 25°C = 25 + 273 = 298K
T2 = 25°C = 298K
Using P1V1/T1 = P2V2/T2
5 × 4/298 = P2 × 2/298
20/298 = 2P2/298
Cross multiply
298 × 20 = 298 × 2P2
5960 = 596P2
P2 = 5960 ÷ 596
P2 = 10 atm.
It will be:
b. Alkyne* series
* You have done a spelling mistake here. It will be "Alkene" not "Alkyne".
Answer:
406.45mL
Explanation:
The following data were obtained from the question:
V1 = 350mL
P1 = 720mmHg
P2 = 630mmHg
V2 =?
The new volume can be obtain as follows:
P1V1 = P2V2
720 x 350 = 620 x v2
Divide both side by 620
V2 = (720 x 350) /620
V2 = 406.45mL
The new volume of the gas is 406.45mL
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.
