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Luba_88 [7]
3 years ago
5

In order for the process to be endothermic, the energy required to break the lattice has to be _______ than the energy released

as the ions go into solution.
a. Greater

b. Less
Chemistry
2 answers:
densk [106]3 years ago
8 0

Explanation:

I honestly believe the answer is greater

REY [17]3 years ago
3 0

Answer:

The answer is Greater

Explanation

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Help please! -A compound contains 0.013 moles of carbon, 0.039 moles of hydrogen and 0.0065 moles of oxygen. Determine the empir
stepan [7]

A) C2H6O1

To find the emperical formula, divide each mole value by the smallest

For carbon, 0.013/0.0065 = 2

For hydrogen, 0.038/0.0065= 6

For oxygen, 0.0065/0.0065= 1

Emperical formula = C2H6O1

8 0
3 years ago
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Rashid [163]

Answer:

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Explanation:

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3 0
3 years ago
in an experiment 3.425g of lead oxide was reduced to form 3.105g of lead the empirical formula of the lead oxide is?​
Romashka-Z-Leto [24]

Answer:

Pb3O4

Explanation:

According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,

Mass of lead oxide = 3.425g

Mass of lead = 3.105g

Mass of oxygen = (3.425g - 3.105g) = 0.320g

Next, we convert each mass value to mole by dividing by respective molar mass

Pb = 3.105g ÷ 207.2 = 0.0149mol

O = 0.320g ÷ 16 = 0.02mol

Next, we divide each mole value by the smallest (0.0149)

Pb = 0.0149mol ÷ 0.0149mol = 1

O = 0.02mol ÷ 0.0149mol = 1.342

Multiply each ratio value by 3 to get:

Pb = 1 × 3 = 3

O = 1.342 × 3 = 4.026

The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.

4 0
2 years ago
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

#SPJ1

7 0
1 year ago
A 4.27 g sample of a lab solution contains 1.81g of acid. What is the concentration of the solution as a mass percentage?
vaieri [72.5K]

Answer:

42.38875878%

Explanation:

i divided 4.27 g from 1.81g using a percentage calculator but im not sure if its correct

5 0
2 years ago
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