Hypothesis, experiment, conclusion<span />
hydrogen and carbon, hope that helped
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:

Explanation:
Hello there!
In this case, according to the given information it turns out possible for us to realize that one mole of the given compound, Mg(ClO₄)₂, has one mole of Mg, two moles of Cl and eight moles of O; thus, we proceed as follows:

Best regards!
Answer:
11.9 is the pOH of a 0.150 M solution of potassium nitrite.
Explanation:
Solution : Given,
Concentration (c) = 0.150 M
Acid dissociation constant = 
The equilibrium reaction for dissociation of
(weak acid) is,

initially conc. c 0 0
At eqm.

First we have to calculate the concentration of value of dissociation constant
.
Formula used :

Now put all the given values in this formula ,we get the value of dissociation constant
.



By solving the terms, we get

No we have to calculate the concentration of hydronium ion or hydrogen ion.
![[H^+]=c\alpha=0.150\times 0.0533=0.007995 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%3D0.150%5Ctimes%200.0533%3D0.007995%20M)
Now we have to calculate the pH.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


pH + pOH = 14
pOH =14 -2.1 = 11.9
Therefore, the pOH of the solution is 11.9