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Schach [20]
2 years ago
11

Help I will mark brainliest

Chemistry
1 answer:
Alja [10]2 years ago
5 0
I think E sorry if I’m wrong
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If the melted lead got hot enough to vaporize, how much energy would it have taken to vaporize 21
arsen [322]

Answer:

18 KJ

Explanation:

Data Given:

mass of Lead (m) = 21 g

Heat taken for vaporization (Q) = ?

Solution:

This problem is related to phase change and latent heat of vaporization.

Latent heat of vaporization is the amount of heat taken to convert one mole of substance at its boiling point to its vapor.

So, Latent heat of vaporization of lead has a constant value

Latent heat of vaporization of lead = 177.7 KJ/mol

Formula used

Q = m x Lv. . . . . (1)

where

Lv = specific latent heat of vaporization

here the value for latent heat of vaporization is for mole so instead of mass we will use moles in formula.

So,

Q = no. of mol x Lv. . . . . (2)

first find no. of moles for 21 g of lead

                  no. of moles = mass in grams / molar mass . . . . . . (3)

molar mass of  lead (Pb) = 207 g/mol

put values in equation 3

                      no. of moles = 21g / 207 g/mol

                      no. of moles = 0.101 mol

so,

number of moles of lead (Pb) = 0.101 mol

Put values in the eq.2

                      Q = 0.101 mol x 177.7 KJ/mol

                      Q = 18 kJ

So, 18KJ of heat is taken to vaporize 21 g of lead (Pb)

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Can anyone tell me about sulfuric acid in shampoo/soap?
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A student is doing an experiment that stoichiometry says should produce 34.6 grams of product. The student actually makes 25.2 g
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which would get you 73%

if you found my answer helpful pls give brainliest
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Has markings along the cylinder/beaker that indicate the volume of liquid inside them. .
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How many oxygen molecules are in 22.4 liters of oxygen gas at 273 K and 101.3 kPa?
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