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.
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Blank 1: nothing (to keep 2 total nitrogen)
blank 2: 3 (to make 6 total hydrogen)
blank 3: 2 (to make 2 total nitrogen and 6 total hydrogen)
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Answer:
a) 25%
b) 27.5 g
c) 90%
Explanation:
a) 75% fat-free by weight means 25% of the weight is made by fat.
b) 110 g ___ 100%
x ___ 25%
x = 27.5g
Each hot dog has 27.5g of fat.
c) 9 cal ___ 1 g fat
y ___ 27.5 g fat
y = 247.5 cal
275 cal ___ 100%
247.5 cal ___ z
z = 90%
90 % of the calories come from fat.
D. dormancy is the correct answer