There are a total of 4 elements
Answer:
There will be produced:
2.97 moles HMnO4
4.45 moles Pb(NO3)2
2.97 moles H2O
Explanation:
Step 1: Data given
Manganese(II) oxide = MnO2
lead(IV) oxide = PbO2
nitric acid = HNO3
Moles of HNO3 = 8.90 moles
Step 2: The balanced equation
2MnO2 + 3PbO2 + 6HNO3 → 2HMnO4 + 3Pb(NO3)2 + 2H2O
Step 3: Calculate moles of reactants and products
For 2 moles MnO2 we need 3 moles PbO2 and 6 moles HNO3 to produce 2 moles HMnO4, 3 moles Pb(NO3)2 and 2 moles of water
For 8.90 moles of HNO3, there will react:
8.90 / 3 = 2.97 moles MnO2
8.90 / 2 = 4.45 moles PbO2
There will be produced:
8.90/3 = 2.97 moles HMnO4
8.90/2 = 4.45 moles Pb(NO3)2
8.90 / 3 = 2.97 moles H2O
Answer:
37.5 mL of a 4.0 M Mg(OH)₂ are needed to make 750 mL of 0.2 M Mg(OH)₂
Explanation:
Dilution is the reduction in concentration of a chemical in a solution. Dilution consists of lowering the amount of solute per unit volume of solution.
In other words, dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one, and consists of adding solvent to an existing solution. The quantity or mass of the solute is not changed but only that of the solvent.
In dilutions the expression is used:
Ci*Vi = Cf*Vf
where:
- Ci: initial concentration
- Vi: initial volume
- Cf: final concentration
- Vf: final volume
In this case:
- Ci: 4 M
- Vi: ?
- Cf: 0.2 M
- Vf: 750 mL
Replacing:
4 M* Vi= 0.2 M* 750 mL
Solving:
Vi= 37.5 mL
<u><em>37.5 mL of a 4.0 M Mg(OH)₂ are needed to make 750 mL of 0.2 M Mg(OH)₂</em></u>
Answer:
The work done by the gas is 117,300 Joules.
Explanation:
Pressure applied on the gas , P= 102.0 kPa = 102000 Pa ( 1kPa = 1000 Pa
Change in volume of the gas ,ΔV= 
Here the work is done by system so, the value of work done will be negative.
The work done by the gas is 117,300 Joules.
Answer:
1) 0.0625 g.
2) 0.0125 g.
Explanation:
<em>1) A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution?</em>
mass% of NaOH = [(mass of NaOH)/(mass of solution)] x 100.
mass% of NaOH = 25.0%, mass of NaOH = ??? g, mass of solution = 0.250 g.
∴ mass of NaOH = (mass% of NaOH)(mass of solution)/100 = (25.0%)(0.250 g)/100 = 0.0625 g.
<em>2) What mass of NaOH must be added to the solution to increase the concentration to 30.00% by mass?</em>
We can use the relation:
mass% of NaOH = [(mass of NaOH)/(mass of solution)] x 100.
mass% of NaOH = 30.0%, mass of NaOH = ??? g, mass of solution = 0.250 g.
∴ mass of NaOH = (mass% of NaOH)(mass of solution)/100 = (30.0%)(0.250 g)/100 = 0.075 g.
∴ The mass of NaOH should be added = 0.075 - 0.0625 = 0.0125 g.
