Which sentence best explains the high melting point, boiling point, and surface tension of water? A. The hydrogen atom on each w
2 answers:
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Answer: 32.94 g
Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:
From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.
for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm
V = 15.0 L
T = 350.8 + 273 = 623.8 K
For krypton,
n = 0.146 moles
for chlorine,
n = 0.439
From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.
Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.
So, the amount of product formed is calculated from moles of krypton.
Molar mass of krypton tetrachloride is 225.61 gram per mol.
There is 1:1 mol ratio between krypton and krypton tetrachloride.
= 32.94 g of
So, 32.94 g of the product will form.
Answer : The correct option is, (A) -101.37 KJ
Explanation :
First we have to calculate the moles of HCl and NaOH.
The balanced chemical reaction will be,
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH
Thus, the number of neutralized moles = 0.1392 mole
Now we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water =
Now we have to calculate the heat absorbed during the reaction.
where,
q = heat absorbed = ?
= specific heat of water =
m = mass of water = 174 g
= final temperature of water = 317.4 K
= initial temperature of metal = 298 K
Now put all the given values in the above formula, we get:
Thus, the heat released during the neutralization = -14.11 KJ
Now we have to calculate the enthalpy of neutralization.
where,
= enthalpy of neutralization = ?
q = heat released = -14.11 KJ
n = number of moles used in neutralization = 0.1392 mole
Therefore, the enthalpy of neutralization is, -101.37 KJ