The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂<span>(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.
the amount of </span>C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions,
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂
Answer:
Explanation: Rutherford model, also called Rutherford atomic model, nuclear atom, or planetary model of the atom, description of the structure of atoms proposed (1911) by the New Zealand-born physicist Ernest Rutherford. The model described the atom as a tiny, dense, positively charged core called a nucleus, in which nearly all the mass is concentrated, around which the light, negative constituents, called electrons, circulate at some distance, much like planets revolving around the Sun. Hope that helps!
Answer:
Na₂CO₃ · 10H₂O
Explanation:
The formula for sodium carbonate hydrate is:
Na₂CO₃ · xH₂O
The unknown "x" is the number of water molecules contained in the hydrate.
To find "x" we have to use the hydrogen percentage in the sample, 7.05 % H.
First we calculate the molecular weight of Na₂CO₃ · xH₂O:
molecular weight of Na₂CO₃ · xH₂O = 23 × 2 + 12 + 16 × 3 + 18x
molecular weight of Na₂CO₃ · xH₂O = 106 + 18x g/mole
Now we devise the fallowing reasoning tanking in account 1 mole of Na₂CO₃ · xH₂O:
if in 106 + 18x grams of Na₂CO₃ · xH₂O we have 2x grams of hydrogen
then in 100 grams of Na₂CO₃ · xH₂O we have 7.05 grams of hydrogen
106 + 18x = (100 × 2x) / 7.05
106 + 18x = 28.4x
106 = 28.4x - 18x
106 = 10.4x
x = 106 / 10.4
x = 10.2 ≈ 10
The formula for the washing soda is Na₂CO₃ · 10H₂O.
We did a lab on this.
hydrogen peroxide on a raw liver bubbles up and stinks.
Answer:
1). 1 mole of Carbon burnt in air
C + O2 →CO2
1 mole of carbon produces 1 mole of CO2 which is 44g of CO2
2). 1 mole of carbon is burnt in 16g of dioxygen
32g of O2 = 44g of CO2
1g of O2 = 44/32
CO2 (Dioxygen is limiting reagent)
16g of O2 = 4/32 × 16 = 22g of CO2 in one mole
3) 2 moles of Carbon burnt in 16g of dioxygen
16g of dioxygen is available, and thus it can combine with 0.5 mol of carbon to give 22g of CO2
