Answer:
37.14 %
Explanation:
Using the equation, mass, M = D1 * V1
= D2 * V2
Where,
D1 = density of the liquid Nitrogen
D2 = density of gaseous Nitrogen
V1 = volume of the liquid Nitrogen
V2 = volume of the gaseous Nitrogen
Calculating V2,
0.808 * 185 = 1.15 * V2
Volume of Nitrogen after expansion = 129.98 m3.
Volume = L * b * h
= 10 * 10 * 3.5
Volume of the room = 350 m3.
Fraction of air = volume of Nitrogen after expansion/volume of the room * 100
= 129.98/350 *100
= 37.14 %
Answer:
21.182 g
Explanation:
There are about (6.0)(10^23) atoms in one mole of a substance, so the given sample has about 0.333 mol of Cu.
The atomic mass of Cu is 63.546 g/mol, meaning that the answer is about <u>21.182</u><u> </u><u>g</u>
Answer:
6 moles of SO₃ formed.
Explanation:
Given data:
Number of moles of SO₃ formed = ?
Number of moles of oxygen react = 3 mol
Solution:
Chemical equation;
2SO₂+ O₂ → 2SO₃
now we will compare the moles of oxygen and sulfur trioxide.
O₂ : SO₃
1 : 2
3 : 2/1×3 = 6 moles
Thus, six moles of SO₃ will formed.
The hydrogen will react with 12 g B
Mass of B = 3.3 g H_2 × (3.6 g B/1.0 g H_2) = 12 g B
Answer: The oxygen gas has a pressure of 4 atm if my volume decreases to 1 L.
Explanation:
Given:
= 2 L,
= 2 atm
= 1 L,
= ?
Formula used to calculate the new pressure is as follows.
![P_{1}V_{1} = P_{2}V_{2}](https://tex.z-dn.net/?f=P_%7B1%7DV_%7B1%7D%20%3D%20P_%7B2%7DV_%7B2%7D)
Substitute the values into above formula as follows.
![P_{1}V_{1} = P_{2}V_{2}\\2 atm \times 2 L = P_{2} \times 1 L\\P_{2} = 4 atm](https://tex.z-dn.net/?f=P_%7B1%7DV_%7B1%7D%20%3D%20P_%7B2%7DV_%7B2%7D%5C%5C2%20atm%20%5Ctimes%202%20L%20%3D%20P_%7B2%7D%20%5Ctimes%201%20L%5C%5CP_%7B2%7D%20%3D%204%20atm)
Thus, we can conclude that the oxygen gas has a pressure of 4 atm if my volume decreases to 1 L.