Question

What volume of 0.115 M HClO4 solution is needed to neutralize 41.00 mL of 8.95×10^−2 M NaOH? 2. What volume of 0.130 M HCl is needed to neutralize 2.79 g of Mg(OH)2? 3. If 27.0 mL of AgNO3 is needed to precipitate all the Cl^− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution? 4. If 45.3 mL of 0.116 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

          What volume of 0.115 M HClO4 solution is needed to neutralize 41.00 mL of 8.95×10^−2 M NaOH?
2. What volume of 0.130 M HCl is needed to neutralize 2.79 g of Mg(OH)2?
3. If 27.0 mL of AgNO3 is needed to precipitate all the Cl^− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution?
4. If 45.3 mL of 0.116 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?
        
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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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What volume of 0.115 M HClO4 solution is needed to neutralize 41.00 mL of 8.95×10^−2 M NaOH? 2. What volume of 0.130 M HCl is needed to neutralize 2.79 g of Mg(OH)2? 3. If 27.0 mL of AgNO3 is needed to precipitate all the Cl^− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution? 4. If 45.3 mL of 0.116 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?
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What volume of 0.115 M HClO4 solution is needed to neutralize 41.00 mL of 8.95×10^−2 M NaOH? 2. What volume of 0.130 M HCl is needed to neutralize 2.79 g of Mg(OH)2? 3. If 27.0 mL of AgNO3 is needed to precipitate all the Cl^− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution? 4. If 45.3 mL of 0.116 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

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Transcript

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00:01 Hello students in this question we have given this data we need to find out volume of hclo4 now here calculate neutralization we have right formula m1 v1 is equal to m2 v2 now we can substitute value we are getting 0 .1 1 1 1 5 into v1 we need to calculate 8 .95 into 10 rise to minus 2 into 41 .00 after calculating we are getting the value of v1 is equal to 31 .910 ml so our final answer is volume of hcl o4 is equal to 31 .910 ml now in that second question we have to find out volume of here we have right the formula we can substitute value we are getting 0 .130 into v1 is equal to n2 is equal to 2 .79 gram divided by 1 .89 gram per mole now after calculating we are getting v1 is equal to 0 .37m.
01:27 So our second answer is volume of volume of volume of scl is equal to 0 .3700ml.
01:38 Now in third question, in that third question, we have to find out molarity of agno3 here.
01:44 We have given this data.
01:45 Now, number of moles of kcl.
01:50 Moles of kcl is equal to 0 .785 into 10 rise to minus 3 gram divided by 74 .5.
02:04 Double 5 is equal to 1 .0530 into 10 x to minus 5 moles...
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