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andre [41]
3 years ago
8

Copper(II) sulfate forms a bright blue hydrate with the formula CuSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high eno

ugh temperature, H 2 O ( g ) can be driven off, leaving the grey‑white anhydrous salt CuSO 4 ( s ) . A 14.220 g sample of the hydrate was heated to 300 ∘ C . The resulting CuSO 4 ( s ) had a mass of 8.9935 g . Calculate the val
Chemistry
1 answer:
Arada [10]3 years ago
5 0

Answer:

The value of n in the hydrate formula is 5 ,CuSO_4.5H_2O.

Explanation:

CuSO_4.nH_2O\rightarrow CuSO_4+nH_2O

Mass of hydrate of copper sulfate = 14.220 g

Moles of hydrate of copper sulfate =\frac{14.220 g}{(159.5+n\times 18) g/mol}

Mass of copper sulfate after heating = 8.9935 g

Moles of copper sulfate = \frac{8.9935 g}{159.5 g/mol}

\frac{14.220 g}{(159.5+n\times 18) g/mol}=\frac{8.9935 g}{159.5 g/mol}

Solving for n, we get:

n = 5

The value of n in the hydrate formula is 5 ,CuSO_4.5H_2O.

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3 years ago
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bixtya [17]

Answer:

C₄H₁₀(g) + O₂(g) ⇒ CO₂(g) + H₂O(g)

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Explanation:

Butane gas (C₄H₁₀) burns in oxygen gas to produce carbon dioxide gas and water vapor. The unbalanced equation is:

C₄H₁₀(g) + O₂(g) ⇒ CO₂(g) + H₂O(g)

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Finally, we will balance the oxygen atoms.

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In order to have integers, we will multiply everý compound by 2.

2 C₄H₁₀(g) + 13 O₂(g) ⇒ 8 CO₂(g) + 10 H₂O(g)

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3 years ago
If i have 17 moles of gas at a temperature of 67c, and a volume of 88.89 liters what is the pressure of the gas
Shtirlitz [24]
Use the state equation for ideal gases: pV = nRT

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V = 88.89 liter
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R = 0.0821 atm * liter / (K*mol)

=> p = nRT / V = 17 mol * 0.0821 (atm*liter / K*mol) * 340.15 K / 88.89 liter

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What is the % by volume of 50mL of ethylene glycol dissolved in 950mL of H2O?
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Answer:

5.0 %

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Step 1: Calculate the volume of solution

If we assume that <em>the volumes are additive</em>, the volume of the solution is equal to the sum of the volume of the solute and the solvent.

V = 50 mL + 950 mL = 1000 mL

Step 2: Calculate the percent by volume

We will use the following expression.

\% v/v = \frac{volume\ of\ solute}{volume\ of\ solution} \times 100 \%  = \frac{50mL}{1000mL} \times 100 \% = 5.0\%

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The oxidation numbers of nitrogen in NH4+, N2O5, and NaNO3 are, respectively:
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2 years ago
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