Answer:
0.4M NaOH
Explanation:
Molarity, M, is an unit of concentration widely used defined as the ratio between moles of solute (In this case, NaOH) and volume of solution in liters.
As the solution contains 2 moles of NaOH-Moles of solute- in 5L of solution, the molarity is:
2 moles NaOH / 5L =
<h3>0.4M NaOH</h3>
It’s ionic
KCl(at)
Potassium chloride
There are three subatomic particles known: (1) electron which is found outside the nucleus of an atom and (2 and 3) protons and neutrons which are both inside the nucleus. As they are outside the nucleus, it is easier to transport electron than any other subatomic particle. Thus, atom and its ion differ in the number of electrons.
Question:
Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reaction of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps:
equations
Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.
Answer:
NO
It is present but not consumed
NO Lowers the activation energy of the reaction
Explanation:
A catalyst is a substance that is present in a chemical reaction and enables the reaction to occur at a faster rte but does not take part n the reaction
Therefore, whereby NO is not consumed, it is the catalyst
It functions by lowering the activation energy
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
