Answer:
1.9 × 10² g NaN₃
1.5 g/L
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles of N₂ formed
N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol
We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.
4.2 mol × 28.01 g/mol = 1.2 × 10² g
Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂
The mass ratio of NaN₃ to N₂ is 130.02:84.03.
1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃
Step 4: Calculate the density of N₂
We will use the following expression.
ρ = P × M / R × T
ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L
For both of them, used the balanced equation and it’s mole ratio to convert whatever you need to into moles. See the attacked work.
1) D 5 mols
2) A 0.55 mols
<span>Jet streams are the major means of transport for weather systems. A jet stream is an area of strong winds ranging from 120-250 mph that can be thousands of miles long, a couple of hundred miles across and a few miles deep. Jet streams usually sit at the boundary between the troposphere and the stratosphere at a level called the tropopause. This means most jet streams are about 6-9 miles off the ground. Figure A is a cross section of a jet stream.
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The dynamics of jet streams are actually quite complicated, so this is a very simplified version of what creates jets. The basic idea that drives jet formation is this: a strong horizontal temperature contrast, like the one between the North Pole and the equator, causes a dramatic increase in horizontal wind speed with height. Therefore, a jet stream forms directly over the center of the strongest area of horizontal temperature difference, or the front. As a general rule, a strong front has a jet stream directly above it that is parallel to it. Figure B shows that jet streams are positioned just below the tropopause (the red lines) and above the fronts, in this case, the boundaries between two circulation cells carrying air of different temperatures.
Answer:
57.6g
Explanation:
So, if in one mole of water, 16 g of oxygen atom is present. Then, in 3.6 moles of water, the mass of oxygen present will be 3.6×16=57.6g. Therefore, the amount of oxygen present in 3.6 g water is option (B)- 57.6 g.
Answer:
2
Explanation:
Each orbital can hold two electrons. One spin-up and one spin-down.
