<u>Answer:</u> The mass of iron in the ore is 10.9 g
<u>Explanation:</u>
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is 
So, in 15.6 g of iron (III) oxide, mass of iron present will be = 
Hence, the mass of iron in the ore is 10.9 g
Answer:
It would be 151.832775 because one mole is 44.0095*3.45 i hope this helps!
Explanation:
The balanced chemical
reaction will be:
CH4 + 2O2 → CO2 + 2H2O
We are given the amount of carbon dioxide to produce from the reaction.
This will be our starting point.
560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) (
22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>
Answer:
D. The rate decreases as reactants are used up.
Explanation:
Initially, the rate increases until the reaction is at equilibrium. At equilibrium, the rate is constant.
As the reaction progresses, the rate decreases to zero when reactants are used up ( for irriversible reactions only )
Answer : The value of rate constant is, 
Explanation :
First we have to calculate the rate constant, we use the formula :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time passed by the sample = 20 min
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 100 - 85 = 15 g
Now put all the given values in above equation, we get
Therefore, the value of rate constant is,
