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4. How many total moles of ions are released when 8.87 moles of Na2HPO4 dissolves completely in water? mol ions 6 How many grams of solute are needed to make 595 mL of 2.58 ×10−2 M potassium sulfate?...

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4.
 How many total moles of ions are released when 8.87 moles of Na2HPO4 dissolves completely in water? mol ions
6
How many grams of solute are needed to make 595 mL of 2.58 ×10−2 M potassium sulfate? _ g solute
8.
If 84.3 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.754 g of precipitate, what is the molarity of lead(II) ion in the original solution?  M
9.
How many milliliters of 0.220 M HCl are needed to react with 22.8 g of CaCO3?
 2HCl(aq) + CaCO3(s) →CaCl2(aq) + CO2(g) + H2O(l)
  ------mL
11.
A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to Fe2+ in acid and then titrating the Fe2+ with MnO4−. A 1.0169− sample was dissolved in acid and then titrated with 28.52 mL of XXXXXXXXXX M KMnO4. The balanced equation is8 H+(aq) + 5 Fe2+(aq)+ MnO4−(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
Calculate the mass percent of iron in the ore.
 -----% Fe 
12
Be sure to answer all parts.
 
The flask (below) represents the products of the titration of 25 mL of sulfuric acid with 25 mL of sodium hydroxide.
 
(a) Write balanced molecular, total ionic, and net ionic equations for the reaction. Include the state of each component in the equations.
 
Molecular equation: _____ -- _______
 
 Total ionic equation (put the reactants on the first line and the products on the second): ____ - ____
Net ionic equation: ____ - ____
13.
 On a lab exam, you have to find the concentrations of the monoprotic (one proton per molecule) acids HA and HB. You are given 42.8 mL of HA solution in one flask. A second flask contains 
37.2 mL of HA, and you add enough HB solution to it to reach a final volume of 50.0 mL.You titrate the first HA solution with 87.3 mL of 0.0906 M NaOHand the mixture of HA and HB in the second flask with 96.4 mL of the NaOH solution. Calculate the molarity of the HA and HB solutions.
 
 M HA
 
 M HB
14.
A typical formulation for window glass is 75.0% SiO2, 15.0% Na2O, and 10.0% CaO by mass. What masses of sand (SiO2), sodium ca
onate, and calcium ca
onate must be combined to produce 0.582 kg of glass after ca
on dioxide is driven off by thermal decomposition of the ca
onates?
 
 kg SiO2
 
 kg Na2CO3
 
 kg CaCO3
15.
In a titration of HNO3, you add a few drops of phenolphthalein indicator to 50.00 mL of acid in a flask. You quickly add 20.00 mL of 0.0930 MNaOH but overshoot the end point, and the solution turns deep pink. Instead of starting over, you add 30.00 mL of the acid, and the solution turns colorless. Then, it takes  7.37 mL of the NaOH to reach the end point. What is the concentration of the HNO3 solution?
Answered 1 days After Sep 23, 2021

Solution

Neelakshi answered on Sep 24 2021
154 Votes
4. How many total moles of ions are released when 8.87 moles of Na2HPO4 dissolves completely in water? mol ions
Answer: 8.87 mol of Na2HPO4
Na2HPO4 = 2Na+ + HPO42-
Na+ = 8.87 mol of Na2HPO4 * 2 mol Na+ / 1mol Na2HPO4
= 1.47 mol Na+
8.87 mol of HPO42- is also present
Total moles of ions = mols of Na+ + mol of HPO42-
= 1.47 + 8.87
= 10.34 mols
6. How many grams of solute are needed to make 595 mL of 2.58 ×10−2 M potassium sulfate?
Answer: Molarity = number of moles of solute in the solvent.
2.58 ×10−2 = moles of solute / 595 mL
Required moles of solute are 15.35 M.
Moles = mass/ molar mass
Mass= Moles* molar mass
Mass = 15.35* 174.29 = 2.675 kg
8. If 84.3 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.754 g of precipitate, what is the molarity of lead(II) ion in the original solution?  M
Answer: Molarity = number of moles of solute in the solvent.
Moles = mass/ molar mass = 0.754* 331.2 = 249.72 M
Molarity = 249.72 /84.3 = 2.96M
9. How many milliliters of 0.220 M HCl are needed to react with 22.8 g of CaCO3?
 2HCl(aq) + CaCO3(s) →CaCl2(aq) + CO2(g) + H2O(l)
Answer: n(HCl)=2⋅n(CaCO3)
The molar mass (MM) of calcium ca
onate is:
MM(CaCO3)=40.1+12.0+3⋅16.0=100.1 gmol
The number of moles of calcium ca
onate:
n(CaCO3)=mass/ MM=22.8/100.1=0.2277 mol
we double the answer to get the amount of HCl needed:
n(HCl)=2⋅n(CaCO3)=2⋅0.2277 =0.4555 mol
Rea
anging the concentration formula
V=n/C
V=0.4555/0.220= 2.0 L
Convert to mL:
V=2.00* 103 mL
11. A chemical engineer determines...
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