Answer: moles
Explanation: Boyle's law:
Boyle's law stated that volume and pressure are inversely proportional to each others by keeping the number of moles and temperature constant.
Mathematical expression:
P₁V₁ = P₂V₂ (n and T are constant)
Charles's law:
Charles's law stated that temperature and volume are directly related related to each others by keeping number of moles and pressure constant.
Mathematical expression:
V₁/T₁ = V₂ /T₂ (n and P are constant)
Gay-Lussac's law:
This law stated that pressure of gas is directly related to its temperature by keeping the number of moles and volume constant.
Mathematical expression:
P₁/T₁ = P₂ /T₂ (n and V are constant)
A solution is a mixture of two or more substances combined so that it is uniform, which means you cannot see the components. <span />
Answer:
About 512 g.
Explanation:
We are given a sample of P₂Cl₅ that contains 179 grams of phosphorus, and we want to determine the grams of chlroine that is present.
Thus, we can convert from grams of phosphorus to moles of phosphorus, moles of phosphorus to moles of chlorine, and moles of chlorine to grams of chlorine.
From the formula, there are two moles of P for every five moles of Cl. The molecular weights of P and Cl are 30.97 g/mol and 35.45 g/mol, respectively. Hence:
In conclusion, there is about 512 grams of chlorine present in the sample.
Alternatively, we can mass percentages. The mass percent of phosphorus in P₂Cl₅ is:
Because there are 179 grams of phosphorus, the total amount of sample present is:
Therefore, the amount of chlorine present is 691.1 g - 179 g, or about 512 g, in agreement with our above answer.
Fe is the limiting reactant.
The balanced chemical equation between iron and oxygen to produce iron (III) oxide is
4Fe(s) + 302(g) - ---> 2Fe2O3(s)
Mass of Fe = 227.8 g
Moles of Fe = 227.8gFe*Imol Fe/55.85g Fe = 4.079mol Fe
Mass of oxygen = 128 g
Moles of O2 = 128g02 * 1molo/32g02 = 4molO2
Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.
Mass of iron (III) oxide produced from Iron = 4.079mol Fe*2molFe2O3/4molFe*159.69g Fe2O3/1 mol Fe2O3 = 325.7gFe2O3
Mass of iron (III) oxide produced from oxygen = 4molO2 *2mol Fe2O3/3molO2*159.69gFe2O3/1mol Fe2O3 = 425.84gFe2O3
Iron (Fe) produces the least amount of the product iron (III) oxide. So, Fe is the limiting reactant.
Learn more about Limiting reagent here:
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