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romanna [79]
2 years ago
15

Someone plssss help meeee !!!! How much energy is required to vaporize 112 g of water? ???

Chemistry
1 answer:
asambeis [7]2 years ago
3 0

Answer:

6.05 × 10⁴ cal

Explanation:

Step 1: Given and required data

Mass of water (m): 112 g

Enthalpy of vaporization of water (ΔHvap): 540 cal/g

Step 2: Calculate how much energy is required to vaporize 112 g of water

Vaporization is a physical change in which a substance goes from the liquid state to the gaseous state. We can calculate the energy required (Q) using the following expression.

Q = ΔHvap × m

Q = 540 cal/g × 112 g = 6.05 × 10⁴ cal

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2 years ago
A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved i
Margaret [11]

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of KNO_3 in original  254.5 mixture.

moles of BaSO_4 = \frac{mass}{Molecular\ Weight}

moles ofBaSO_4  = \frac{68.3}{233.38}

                               = 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of BaCl_2 = mol\times molecular weight

                         = 0.2926\times 208.23

                         = 60.92 g

the AgCl moles = \frac{mass}{Molecular\ Weight}

                          = \frac{199.1}{143.32}

                          = 1.3891 mol of AgCl

note that, the Cl- derive from both, BACl_2 and NaCl

so

mole of Cl- f NaCl = (1.3891) - (0.2926\times 2) = 0.8039 mol of Cl-

mol of NaCl = 0.8039 moles

mass = mol\times Molecular\ Weight  = 0.8039 \times 58 = 46.626\ g \ of \ NaCl

then

KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3

Mass of KNO3 in the original mix is 146.954 g

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