Tropical or warm air masses form in the tropics and have low air pressure.
polar or cold air masses form north of 50 degrees north latitude and south of 50 degrees south latitude
Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
Answer:
3,29L
Explanation:
3.29L = V2
Formula: V1/T1 = V2/T2
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Given:
V1 = 3.0 L V2 = ?
T1 = 310 K T2 = 340 K
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Plugin:
(X stands in place of V2 just to make it easier to look at)
[3.0L / 310K = X / 340K]
(3.0L / 310K = 0.01L/K)
0.01L/K = X / 340K
(multiply 340K on both sides, it cancels out on the right)
0.01L/K * 340K = X
(0.01L/K * 340K = 3.29L)
**3.29L = X**
[or]
**3.29L = V2**
Wax is definitely susceptible to heat and with the application of heat such as burning a candle, the solid wax gets converted into a liquid. This fact can be used to remove ear wax also actually using hollow candles that will melt the ear wax and allow it to be drained out of the ear.
