Answer A: Connect a wire coil to an ammeter. Move a bar magnet into and out of the wire coil as you observe the ammeter.
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
Answer: Ammonia (NH3) and sodium carbonate (Na2CO3), because they accept hydrogen ions but lack hydroxide ions.
Explanation:
i took the test and got it correct :) hope this helps
Answer:
it increases and is perpendicular to the motion of the wave.
Answer:
13.2 g Na2CO3
Explanation:
Convert 10.0 g NaOH to mol.
10.0 g x 1 mol/39.997 g = 0.250 mol
Use mol ration given by the equation: 2 mol NaOH to 1 mol Na2CO3
0.250 mol NaOH x 1 mol Na2CO3/2 mol NaOH = 0.125 mol Na2CO3
Finally, convert the moles of Na2CO3 to grams.
0.125 mol Na2CO3 x 105.99 g/1 mol = 13.2 g