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3 years ago

9

A 270.0 gram piece of copper is dropped into 300.0 grams of water 24.00 °C. If the final temperature of water is 42.00 °C, what

was the initial temperature of the copper piece?
Specific heat of copper = 0.39 J/g °C

257 °C
272 °C
302 °C
313 °C

1 answer:

Papessa [141]3 years ago

4 0

Answer:

257

Explanation:

Just finished the test

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In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

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Where:

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For <em>hydrogen </em>we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for <em>nitrogen</em>:

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$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

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$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

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Answer:

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