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Agata [3.3K]
3 years ago
5

Calculate the molarity of 22.5 g of MgS in 829 mL of solution

Chemistry
1 answer:
krek1111 [17]3 years ago
5 0

Answer:18652.5

Explanation

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What is the IUPAC name of the following compounds i. CO₂ ii. NO₂ iii. HNO3 iv. H₂SO4 v. K-Cr₂O vi. [Fe(CN)
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Explanation:

I) carbon dioxide

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A person walks north for 5 miles, then south for 8 miles, and then north for 6 more miles. The entire trip takes 8 hours.
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C12H7Cl3FNaO2 what are them elements in that chemical formula
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8 0
3 years ago
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
Which organism is a producer? A:owl B:mouse C:grass D:snake please somebody help me
ArbitrLikvidat [17]

Answer:

I think it's C. Grass. :)

7 0
3 years ago
Read 2 more answers
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