Mass of KCl= 19.57 g
<h3>Further explanation</h3>
Given
12.6 g of Oxygen
Required
mass of KCl
Solution
Reaction
2KClO3 ⇒ 2KCl + 3O2
mol O2 :
= mass : MW
= 12.6 : 32 g/mol
= 0.39375
From the equation, mol KCl :
= 2/3 x mol O2
= 2/3 x 0.39375
=0.2625
Mass KCl :
= mol x MW
= 0.2625 x 74,5513 g/mol
= 19.57 g
I remember coming across this question and the options were:
KOH, HCN, NH₃, HI, Sr(OH)₂
Now, a substance with a low pH is one that dissociates completely in water to release hydrogen ions, while basic substances dissociate completely to release hydroxide ions. Therefore, in the order of increasing pH:
HI, HCN, NH₃, Sr(OH)₂, KOH
The average atomic mass of an element can be determined by multiplying the individual masses of the isotopes with their respective relative abundances, and adding them.
Average atomic mass of Br = 158 amu(0.2569) + 160 amu(0.4999) + 162 amu(0.2431)
Average atomic mass = 159.96 amu
As described in the problem, the relative abundance for Br-79 is 25.69%. This is because 2 atoms of Br is equal to 79*2 = 158 amu. Similarly, the relative abundance of Br-81 is 81*2 = 162, which is 24.31%.
Answer: 5.0 moles
Explanation:
From the equation, we see that for every 4 moles of ammonia consumed, 4 moles of nitrogen monoxide are produced (we can reduce this to moles of ammonia consumed = moles of nitrogen monoxide produced).
This means that the answer is <u>5.0 mol</u>
<u>Answer:</u> The net ionic equation is written below.
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of sodium carbonate and nickel (II) chloride is given as:
Ionic form of the above equation follows:
As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
Hence, the net ionic equation is written above.
