The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL
<u>Given data :</u>
Concentration of siilver nitrate ( M₁ ) = 6.0 M
volume of solution ( V₁ ) = 500 mL
Conc of solution ( M₂ )= 1.2 M
<h3>Determine the amount of water that must be added</h3>
we will apply the equation below
M₁V₁ = M₂V₂ ---- ( 1 )
where : V₂ = V₁ + water added ---- ( 2 )
V₂ ( Final volume ) = ( M₁V₁ ) / M₂
= ( 6 * 500 ) / 1.2
= 2500 mL
Back to eqaution ( 2 )
2500 mL = 500 mL + added water
therefore ; added water = 2500 - 500
= 2000 mL
Hence we can conclude that The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL.
Learn more about Volume : brainly.com/question/12410983
#SPJ1
Answer:
C because 77.0 CsF
Explanation:
That is the correct answer because I have that homework and i got it right
By adding the enthalpies of the intermediate reactions together to get the enthalpy of the desired reaction
633.97 L
Explanation:
Well use the combined gas law;
P₁V₁T₁ = P₂V₂T₂
We need to change the temperatures into Kelvin;
18.9°C= 292.05 K
5.9°C = 279.05 K
756 * 512 * 292.05 = 639 * V₂ * 279.05
113,044,377.6 = 178,312.95 V₂
V₂ = 113,044,377.6 / 178,312.95
V₂ = 633.97 L
Answer:
Write this in a word and skeleton equation:
Solid silver chloride and an aqueous solution of nitric acid are produced when a solution of silver nitrate is reacted with a solution of hydrochloric acid.
Explanation:
