Answer:
3.18 L
Explanation:
Step 1: Given data
- Initial pressure (P₁): 0.985 atm
- Initial volume (V₁): 3.65 L
- Final pressure (P₂): 861.0 mmHg
Step 2: Convert P₁ to mmHg
We will use the conversion factor 1 atm = 760 mmHg.
0.985 atm × 760 mmHg/1 atm = 749 mmHg
Step 3: Calculate the final volume of the gas
Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L
Explanation:
Element X forms a chloride with the formula XCl 2 , which is a solid with a high melting point. X would most likely be in the same group of the Periodic Table as. (a) Na (b) Mg (c) Al (d) Si. the answer is Mg .
Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles =
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O =
= 0.13moles
Number of moles of NH₃ =
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
The runner ran a total of 8800 yards.
<span> Greenhouse gases were not historically present in the atmosphere.</span>