coefficient: they balance the chemical equation you have to make sure the number is as small as it can. It is also used to convert different compounds to compounds or quantities to quantities.
The volume of a sample of ammonia gas : 5.152 L
<h3>Further explanation</h3>
Given
0.23 moles of ammonia
Required
The volume of a sample
Solution
Assumed on STP
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So for 0.23 moles :
= 0.23 x 22.4 L
= 5.152 L
Answer:
1) 0.18106 M is the molarity of the resulting solution.
2) 0.823 Molar is the molarity of the solution.
Explanation:
1) Volume of stock solution = 
Concentration of stock solution = 
Volume of stock solution after dilution = 
Concentration of stock solution after dilution = 
( dilution )
0.18106 M is the molarity of the resulting solution.
2)
Molarity of the solution is the moles of compound in 1 Liter solutions.
Mass of potassium permanganate = 13.0 g
Molar mass of potassium permangante = 158 g/mol
Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)
0.823 Molar is the molarity of the solution.
Answer:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
E decreseas 3/2 as fast as G increases = 0.30 M/s
Explanation:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:
Given data = d[D]/dt = 0.10 M/s
-d[D] / 2dt = d[H]/dt
d[H]/dt = 0.05 M/s
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:
d[G] / 2dt = -d[H]/3dt
E decreseas 3/2 as fast as G increases = 0.30 M/s
i think its C pls dont get mad if it is wrong
