Answer:
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I'm not exactly sure which one but I do know that an acid and a base react in a aqueous solution to form water, so i would probably eliminate the ones that aren't aqueous solutions.
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
Explanation:
Bond order is inversely proportional to the bond length.

In
molecule. one nitrogen is double bonded to nitrogen and one oxygen is single bonded to nitrogen and hydrogen bond.
- Bond order between the (N=O) bond is 2 which means that bond length between the (N=O) bond is shorter than that of the N-O bond.
- Bond order between the (N-O) bond is 1 which means that bond length of the N-O bond is longer than that of the bond length of (N=O) bond.
Answer: Volume of gas in the stomach, V = 0.0318L or 31.8mL
Explanation:
The number of moles of oxygen will remain constant even though the liquid oxygen will undergo a change of state to gaseous inside the person's stomach due to an increase in temperature.
<em>Number of moles of oxygen gas = mass/molar mass</em>
molar mass of oxygen gas = 32 g/mol
mass of oxygen gas = density * volume
mass of oxygen gas = 1.149 g/ml * 0.035 ml
mass of oxygen gas = 0.040215 g
Number of moles of oxygen gas = 0.0402 g/(32 g/mol)
Number of moles of oxygen gas = 0.00125 moles
<em>Using the ideal gas equation, PV=nRT</em>
where P = 1.0 atm, V = ?, n = 0.00125 moles, R = 0.082 L*atm/K*mol, T = (37 + 273)K = 310 K
<em>V = nRT/P</em>
V = (0.00125moles) * (0.082 L*atm/K*mol) * (310 K) / 1 atm
V = 0.0318L or 31.8mL