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icang [17]
3 years ago
10

In an experiment, a student gently heated a hydrated copper compound to remove the water of hydration. the following data was re

corded:
1. mass of the crucible, cover and contents before heating 23.4g.

2. mass of empty crucible and cover.

3. mass of crucible, cover, and contents after heating to constant mass 20.94g
Chemistry
1 answer:
NeX [460]3 years ago
3 0
The unknown of this problem is the experimental percent of water in the compound in order to remove the water of hydrogen, given the following:

Mass of crucible, cover and contents before heating                               23.54 g
Mass of empty crucible and cover                                                            18.82 g
Mass of crucible, cover, and contents after heating to constant mass     20.94 g

In order to get the answer, determine the following:

Mass of hydrated salt used                          = 23.54 g – 18.82 g = 4.72 g
Mass of dehydrated salt after heating          = 20.94 g – 18.82 g = 2.12 g
Mass of water liberated from salt                 = 4.72 g – 2.12 g = 2.60 g

Then solve the percent of water in the hydrated salt by:

% water = (mass of water / mass of hydrated salt) x 100
% water = 2.60 g / 4.72 g x 100
% water = 55.08 % in the compound 
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Review the PCR (Calvin-Benson-Basham) cycle and the photochemical (light) reactions. If photosynthesizing green algae are provid
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Explanation:

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Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).
nalin [4]

The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE  for C≡O  = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

To learn more about enthalpy here

brainly.com/question/13981382

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