The atomic number of an element is based on the number of protons in the atomic nuclei of its atoms.
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:
K I will attempt
Explanation:
a)
b)
1 : 2 : 2 (I don't know if this is what the question wants but it is what I would answer)
c)
Hydrogen because it requires 2 moles of H2 to react with 1 mole of O2
d)
24 moles of water. Look at stoichiometric coefficient. 2:2 means 24 moles you get 24 moles
e)
Oxygen. 2 < 5/2. Remember, 1 mole of O2 requires 2 moles of H2. But 5/2 is still greater than 2
f)
First, let's find out how many moles of water we can get. Since O2 is the limiting reactant, and O2:H2O ratio is 1:2, we will get 4 moles of H2O. Then, we can multiply 4 by Avogadro's number which is 
to get the number of molecules. We get: 2.41 * 10^24 molecules of water.
The rate of movement increases, as they get faster with more energy.
Answer:
true!
Explanation:
if youre going by the scientific method, observations are mainly first. then questions, research, etc
