Answer:
1.02 × 10⁶ g
Explanation:
Step 1: Given data
- Volume of the balloon (V): 5400 m³
- Absolute pressure (P): 1.10 × 10⁵ Pa
- Molar mass of He (M): 4.002 g/mol
Step 2: Convert "V" to L
We will use the conversion factor 1 m³ = 1000 L.
5400 m³ × 1000 L/1 m³ = 5.400 × 10⁶ L
Step 3: Convert "P" to atm
We will use the conversion factor 1 atm = 101325 Pa.
1.10 × 10⁵ Pa × 1 atm / 101325 Pa = 1.09 atm
Step 4: Calculate the moles of He (n)
We will use the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.09 atm × 5.400 × 10⁶ L / 0.08206 atm.L/mol.K × 280 K
n = 2.56 × 10⁵ mol
Step 5: Calculate the mass of He (m)
We will use the following expression.
m = n × M
m = 2.56 × 10⁵ mol × 4.002 g/mol
m = 1.02 × 10⁶ g
Answer:
No because it is stayed that way and you can't define them differently.
<span>1. The two qualities used to describe winds are direction and speed.
2. a local wind that blows during the day from an ocean toward land is a(n) sea breeze.
3. The increase in cooling that wind can cause is called the wind-chill factor.
4. Temperature differences between the equator and poles produce convection currents.
A movement that is parallel to Earth's Surface is called wind and a local wind is that wind that blows over a short distance.</span>
Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
Answer:
Far right
Explanation:
The qualities indicated here are of the element Sulphur (S) and any element of same qualities i.e color pale yellow, nature solid, electrical conductivity none, should be placed at far right corner of the periodic table or position 3 mentioned in the attached picture.
Hope it help!
