A heat energy of 645 J is applied to a sample of glass with a mass of 28.4 g. Its temperature increases from –11.6 ∞C to 15.5 ∞C
1 answer:
You might be interested in
Answer:
6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
Explanation:
We are given the chemical equation:
And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.
Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:
- The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
- The ratio between NH₃ and Cu is 2:3.
- The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)
Dimensional Analysis:
- The amount of N₂ produced:
- The amount of Cu produced:
- And the amount of H₂O produced:
In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
Answer: The empirical formula for the given compound is
Explanation:
The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:
where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.
We are given:
Mass of
Mass of
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 21363 g of carbon dioxide, of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in 6125 g of water, of hydrogen will be contained.
Now we have to calculate the mass of nitrogen.
Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Nitrogen =
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.
For Carbon =
For Hydrogen =
For Nitrogen =
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 5 : 7 : 1
Hence, the empirical formula for the given compound nicotine is