They are the tides and conditions of the oceans
Answer:+3
Explanation: You have to find the oxidation number of N which is +3.
Step 1: Change density from g/mL to g/L;
0.807 g/mL = 807 g/L
Step 2: Find Moles of N₂;
As,
Density = Mass / Volume
Or,
Mass = Density × Volume
Putting Values,
Mass = 807 g/L × 1 L
Mass = 807 g
Also,
Moles = Mass / M.mass
Putting values,
Moles = 807 g / 28 g.mol⁻¹
Moles = 28.82 moles
Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,
As,
P V = n R T
V = n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)
V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm
V = 704.76 L
Answer:
Not doubled
Explanation:
The equation below represent the ideal gases relationship
PV ÷ T = constant
Here
P denotes pressure,
V denotes volume,
T denotes temperature in degrees Kelvin
Now
20 ° c = 273 + 20
= 293 K
And,
40 ° c = 313 K
So,
V = Vo. 313 K ÷ 293 K = 1.07 Vo
So, the volume is NOT doubled.
In the case when the temperature would be determined in degrees celsius at 0 degrees so the volume would be zero
