6 L of oxygen (O₂)
Explanation:
We have the following chemical reaction:
2 Mg + O₂ → 2 MgO
Now we calculate the number of moles of magnesium:
number of moles = mass / molar weight
number of moles of Mg = 12 / 24 = 0.5 moles
Taking in account the chemical reaction we devise the following reasoning:
if 2 moles of Mg react with 1 mole of O₂
then 0.5 moles of Mg react with X moles of O₂
X = (0.5 × 1) / 2 = 0.25 mole of O₂
mass = number of moles × molar weight
mass of O₂ = 0.25 × 32 = 8 g
Knowing the information given by the problem that 32 g of O₂ takes up 24 liters of space, we devise the following reasoning:
if 32 g of O₂ takes up 24 liters of space
then 8 g of O₂ takes up Y liters of space
Y = (8 × 24) / 32 = 6 L of O₂
Learn more about:
balancing chemical reactions
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As you move across a period, the atomic radii decreases. ... As you move across a period, electrons are added to the same energy level while protons are also being added. The concentration of more protons creates a higher effective nuclear charge.
Answer:
The theoretical yield of Cu(s) in moles is 60.15 moles
Explanation:
Step 1: Data given
Number of moles CuO = 70.8 moles
Number of moles NH3 = 40.1 moles
Molar mass CuO = 79.545 g/mol
Molar mass NH3 = 17.03 g/mol
Step 2: The balanced equation
3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)
For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2
NH3 is the limiting reactant. It will completely be consumed (40.1 moles). CuO is in excess. There will react 3/2 * 40.1 = 60.15 moles
There will remain 70.8 - 60.15 = 10.65 moles CuO
Step 3: Calculate moles Cu
For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2
For 40.1 moles NH3 we'll have 60.15 moles Cu
The theoretical yield of Cu(s) in moles is 60.15 moles
Applied force
Gravitational force
And Normal force
