Please show the full question!
Answer:
0.175mol
Explanation:
Molarity of a solution can be calculated thus:
Molarity (M) = number of moles (n) / volume (V)
According to the information provided in this question, volume of NH4OH = 0.125L, molarity = 1.4M
Using; Molarity = n/V
1.4 = n/0.125
n = 0.175mol
Answer:
The answer to your question is:
Explanation:
Data
carbon 7.3% = 7.3g
hydrogen 4.5% = 4.5g
oxygen 36.4% = 36.4 g
nitrogen 31.8% = 31.8 g
Now
For carbon
12 g --------------------1 mol
7.3 g ------------- x
x = 7.3/12 = 0.608 mol
For hydrogen
1 g -------------------- 1 mol
4.5 g ------------------ x
x = 4.5 mol
For oxygen
16 g ------------------- 1 mol
36.4 g ---------------- x
x = 2.28 mol
For nitrogen
14 g ---------------- 1 mol
31.8 g --------------- x
x = 2.27 mol
Now divide by the lowest result, the is 0.608 from carbon
carbon 0.608/0.608 = 1
hydrogen 4.5/ 0.608 = 7.4
oxygen 2.28/0.608 = 3.75
nitrogen 2.27/0.608 = 3.73
Empirical formula = CH₇O₄N₄
Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Answer :
(a) The density of mercury is, 13.6 g/ml
(b) The mass of 120.0 ml of mercury is, 1632 grams
Explanation :
(a) Now we have to calculate the density of mercury.
<u>Given :</u>
Volume of mercury = 25.0 ml
Mass of mercury = 340.0 g
Formula used :
Therefore, the density of mercury is, 13.6 g/ml
(b) Now we have to calculate the mass of 120.0 ml of mercury.
As, 25.0 ml of mercury has mass = 340.0 g
So, 120.0 ml of mercury has mass = 
Therefore, the mass of 120.0 ml of mercury is, 1632 grams
