Answer:
Sodium is extracted from it's ore by electrolysis of fused sodium chloride.
Explanation:
he process is usually carried out ia a special electrochemical cell called the downs cell. While molten sodium metal is collected at the cathode and also sent to tanks for cooling and storage.
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
Answer:
Explanation:
Mass of F / Mass of S = 2.962/1 =2.962 X 32 / 32 = 94.78/32
Mass of F / Mass of S = 2.370 /1 = 2.370 X 32 / 32 = 75.84 /32
Mass of F /Mass of S = 3.555/1 = 3.555 x 32 / 32 = 113.76 / 32 .
Now constant mass of S that is 32 g reacts with different mass of F. They are as follows :
94.78 g , 75.84 g , and 113.76 g
Their ratio = 94.78 : 75.84 : 113.76
divide them by 19
their ratio = 5 : 4 : 6
So these data are consistent with law of multiple proportion.
Must contain: 6 protons, 6 electrons and 12 neutrons.
NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>