First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
Answer:
4.8 h
Explanation:
The meal contains 53.0 grams of fat. If fat provides 38 kJ/g, the energy supplied is:
53.0 g × 38 kJ/g = 2.0 × 10³ kJ
The meal contains 38.0 grams of protein. If protein provides 17 kJ/g, the energy supplied is:
38.0 g × 17 kJ/g = 6.5 × 10² kJ
The meal contains 152 grams of carbohydrate. If carbohydrate provides 17 kJ/g, the energy supplied is:
152 g × 17 kJ/g = 2.6 × 10³ kJ
The total energy supplied is:
2.0 × 10³ kJ + 6.5 × 10² kJ + 2.6 × 10³ kJ = 5.3 × 10³ kJ
Swimming burns 1100.0 kJ/hour. The time required to burn 5.3 × 10³ kJ is:
5.3 × 10³ kJ × (1 h/1100.0 kJ) = 4.8 h
Answer:
The correct option is the last option
Explanation:
The chemical reaction provided in the question is a double-displacement reaction which is an exothermic reaction (which is the reason for the release of heat). An example of a double displacement reaction and exothermic reaction is the neutralization reaction illustrated below.
HCl + NaOH ⇒ NaCl + H₂O
From the law of conservation of energy, energy can neither be created nor destroyed, hence the total energy in a given system should ordinarily be the same (in the reactants and products), however <u>when energy is released in a reaction (as in the case with an exothermic reaction), it shows there are more bond energy in the reactants than in the products and it is the excess energy that is been released into the atmosphere.</u>
Answer:
There will be no observed impact of adding twice as much Na2CO3 on the product
Explanation:
Stoichiometry gives the relationship between reactants and products in terms of mass, mole and volume.
If we consider the stoichiometry of the reaction, we will discover that the reaction occurs in a 1:1 ratio. This implies that use of twice the amount of Na2CO3 will only lead to an excess of Na2CO3 making the other reactant the limiting reactant. Once the other reactant is used up, the reaction quenches.
Hence, use of twice as much Na2CO3 has no impact on the quantity of product produced.
