Answer: depending on the method, 138-139 (imperial) gallons, or 626 L
Explanation:
The wt of water is 4924-3547 lb = 1377 lb = 1377/2.2 kg = 626 kg = 626 L
1gallon = 4.5 L (it does where I come from and who still measures things in imperial anyway?) so I guess that is 139 gallons
or alternatively, I recall from distant childhood, “a pint of water weighs a pound and a quarter” which means 1 gallon = 10 lb, so 1377 lb = ~138 gallons
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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