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Alecsey [184]
3 years ago
7

A 10.0 g sample of a hydrate was heated until all the water was driven off. The mass of the anhydrous product remaining was 8.00

grams. What is the percent of water in the hydrate?
Chemistry
1 answer:
chubhunter [2.5K]3 years ago
3 0

Answer:

20%

Explanation:

10.0 = 100%

8.00 = 80%

100-80= 20

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Chemists can use moles to calculate: A. How much of the products are needed and how much reactant will be made. B. How much prod
damaskus [11]

Answer:

c.- How much of the reactants are needed and how much product will made.

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Enthalpy of <br><br> CH4(g) + 2NO2(g) -&gt; N2(g) + CO2(g) + 2H2O(l)
stira [4]

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-177.9 kJ.

Explanation:

Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.

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2 years ago
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