Answer:According to the data results and the period table, these elements/substances are similar because they are in the same group; not only that but they both discharge from metals.
Explanation: I got you
Data:
V1 = 6.7 liter
T1 = 23° = 23 + 273.15 K = 300.15 K
P1 = 0.98 atm
V2 = 2.7 liter
T2 = 125° = 125 + 273.15 K = 398.15 K
P2 = ?
Formula:
Combined law of ideal gases: P1 V1 / T1 = P2 V2 / T2
=> P2 = P1 V1 T2 / (T1 V2)
P2 = 0.98 atm * 6.7 liter * 398.15 K / (300.15K * 2.7 liter)
P2 = 3.22 atm
Answer:
10.80
Explanation:
As per the equation, let us calculate the mole ratio. N2+3H2→2NH3. As per the equation one mole of nitrogen reacts with 1 mol of hydrogen.
In terms of mass. 28.01 g of nitrogen needs 3 mol of hydrogen or 6.048 g of hydrogen.
We can set up the ratio;
28.01 g of
l
N
2
needs
6.048 g of
l
H
2
1 g of
l
N
2
needs
6.048
28.01
g of
l
H
2
50.0 g of
l
N
2
needs
6.048
×
50.0
28.01
l
g of
l
H
2
=
10.80 g of
l
H
2
