Answer:
A decrease in temperature would decrease kinetic energy, therefore decreasing collisions possible.
Explanation:
A gas at a fixed volume is going to have collisions automatically. If you decrease the temperature (same thing as decreasing kinetic energy) you are cooling down the molecules in the container which gives them less energy and "relaxes" them. This decrease in energy causes them to move around much slower and causing less collisions, at a much slower rate. In a perfect world, these collisions do not slow down the molecule but we know that they do, just a very very small unmeasurable amount.
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.
The formula of alcohol are
1.913 g CO2/44 = 0.043 moles of CO2*12 = 0.043 moles of C 1.174 gram H2O/18 = 0.065 moles of H.
O: 1 - 0.516 g C - 13 g H = 0.354 g O/16 = 0.022 moles of O.
we have 0.043:0.130.;0.022 as C:H:O. Diving by the smallest no. 0.022 we get 1.95:5.9:1 or 2:6:1 as C:H:O.
hope this help
Answer:
The uneven heating of Earth's surface
Explanation: