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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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a - Increasing the rate of the forward reaction will cause equilibrium to be more product-favored (i.e. shift to the right, not the left)

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3 years ago
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2L of hydrogen has an initial pressure of 750 mmHg, what is the final pressure in mmHg if the volume increases to 20 L with a co
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Answer: The final pressure is 75 mm Hg.

Explanation:

According to Boyle's law, at constant temperature the pressure of a gas in inversely proportional to volume.

Since, it is given that the temperature is constant. Hence, formula used is as follows.

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Substitute the values into above formula as follows.

P_{1}V_{1} = P_{2}V_{2}\\750 mm Hg \times 2 L = P_{2} \times 20 L\\P_{2} = \frac{750 mm Hg \times 2 L}{20 L}\\= 75 mm Hg

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3 0
3 years ago
A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.
BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\  &=185.59\text{ g/mol} \end{aligned}

Dimensional Analysis:

\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

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