12) How many milliliters (mL) of a 3.00 M solution of sodium chlorate must be ad
ID: 1082762 • Letter: 1
Question
12) How many milliliters (mL) of a 3.00 M solution of sodium chlorate must be added to 275.0 milliliters (mL) of water to achieve a 0.575 M sodium chlorate solution? (3) Use dimensional analysis for the following three questions, show your work 13) Consider the following reaction: Pb(NO3)2(a)2Nal(a) 2NaNOsa)Pbl2(e) How many mL of a 0.100 M solution of lead nitrate is needed to completely react with 75.0 mL of a 0.200 M solution of sodium iodide? (6) 14) Consider the following reaction: 6 HCl(aq) + 2 Al(s) 2 AICI3(aq) + 3 H2 How many grams of aluminum are needed to completely react with 35.0 mL of a 2.00 M solution of hydrochloric acid? (5) the following reactions Ca(NO3)2(aNa2COs(aa) CaCOs() 2 NaNOs(ag) If 148.0 grams of sodium carbonate react, how many milliliters (mL) of a 0.750 M solution of calcium nitrate are needed? (5)Explanation / Answer
Answer 13)
let the volue of water is x ml
by using the dilution law M1V1=M2V2
M1 = initial concentration= 3 M
V1= initial volume = x mL
M2= final concentration = 0.575 M
V2= final volume = volume of sodium chlorate +volume of water = (x+275)mL
using the values
3*x= 0.575*(x+275)
3x=0.575x+ 158.125
3x-0.575x= 158.125
2.425x=158.125
x= 158.125/2.425 = 65.20 mL
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