1.) Mass
2.) Can occupy space (Volume)
Answer:
Volume of ammonia produced = 398.7 dm³
Explanation:
Given data:
Volume of N₂ = 200 dm³
Pressure and temperature = standard
Volume of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of N₂:
PV = nRT
1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K
n = 200 atm.L /22.41 atm.L/mol
n = 8.9 mol
Now we will compare the moles of ammonia and nitrogen.
N₂ : NH₃
1 : 2
8.9 : 2/1×8.9 = 17.8 mol
Volume of ammonia:
1 mole of any gas occupy 22.4 dm³ volume
17.8 mol ×22.4 dm³/1 mol = 398.7 dm³
Explanation:
there is 2 nitrogen but if you mean nitrate is 6
Answer:
Carbon dioxide levels in the Earth's atmosphere have been steadily increasing.
Carbon has a longer average lifetime in the atmosphere.
Explanation:
Today the level of carbon dioxide is higher than at any time in human history. Scientists widely agree that Earth’s average surface temperature has already increased by about 2 F (1 C) since the 1880s, and that human-caused increases in carbon dioxide and other heat-trapping gases are extremely likely to be responsible.
The lifetime in the air of CO2, the most significant man-made greenhouse gas, is probably the most difficult to determine, because there are several processes that remove carbon dioxide from the atmosphere. Between 65% and 80% of CO2 released into the air dissolves into the ocean over a period of 20–200 years.
<span>Use the van't Hoff equation:
ln
(
K2
K1
)
=
Δ
HÂş
R
(
1
T1
â’
1
T2
)
ln
(
K2
7.6*10^-3
)
=
-14,200 J
8.314
(
1
298
â’
1
333
)
ln
(
K2
7.6*10^-3
)
=
â’
1708
(
0.00035
)
ln
(
K2
0.0076
)
=
â’
0.598
Apply log rule
a
=
log
b
b
a
-0.598 =
ln
(
e
â’
0.598
)
=
ln
(
1
e
0.598
)
Multiply both sides with e^0.598
K
2
e
0.598
= 0.0076
K
e
0.598
e
0.598
=
0.0076
e
0.598
K
2
=
0.0076
e
0.598
=
4.2
â‹…
10
â’
3
K2
=
4.2
â‹…
10
â’
3</span>
