Answer:
I needed to use good precision in my measurement for my chemistry lab
Explanation:
40.1g of nitrogen gas is produced.
The equation given is
2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O
This equation is already balanced.
When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.
We get 1 mole of nitrogen from 3 moles of copper oxide.
We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.
4.3/3 x 1 = 1.433 mols
- 1.433 mols of nitrogen gas are produced
- The molar mass of nitrogen gas is 14+14 = 28g
- The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g
40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.
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Answer:
C. Digestion is the answer.
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
Answer: B. Allow light to pass through. :)
