Answer:
The new concentration is 2.03M
Explanation:
Step 1: Data given
A 200 mL 3.55 M HBr is diluted with 150 mL
Step 2: The dilution
In a dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution equals the ratio that exists between the volume of the diluted solution and the volume of the stock solution.
Dilution factor = [stock sample]/[diluted sample] = diluted volume / stock volume
In this case, the volume of the stock solution is 200 mL
Adding 150 mL of water to the stock solution will dilute it to a final volume of 200 + 150 = 350 mL
The dilution factor wll be 350/200 = 1.75
This makes the diluted concentration:
3.55/1.75 = 2.03M
The new concentration is 2.03M
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Hope this helps!
B. More n2and I wwould form
For the whole set of problems, always remember the Avogadro’s
number is 6.023*10^23 units per mole of a substance. Units could be atoms,
molecules or formula units.
The first question asks for the number of molecules of NaNO3.
The molar mass of NaNO3 is 85 grams per mole. So,
<span>150g NaNO3(1mole NaNO3/85 grams NaNO3)(6.023*10^23
molecules/1mole NaNO3)=1.063*10^24 molecules of NaNO3</span>
<span>5.7*10^46 molecules of NaNO3(1mole NaNO3/6.023*10^23
molecules)(85 grams NaNO3/1mole NaNO3) = 8.044*10^24 grams NaNO3</span>
For the molar mass of water, we have 18.02grams per mole.
301 moles H2O(18.02 grams H2O/1
mole H2O) = 5424.02 grams H2O
For the molar mass of sulfuric
acid, we have 98.08 grams per mole.
25g H2SO4(I mole H2SO4/98.08g H2SO4) =
0.2549 mole H2SO4
For the molar mass of Ca(OH)2,
we have 74.1 grams per mole.
252gCa(OH)2(1mol/74.1g)(6.023*10^23/1mol)
= 2.048*10^24 molecules of Ca(OH)2
For the molar mass of calcium,
we have 40 grams of Ca per mole.
6.7*10^35 atoms Ca(1 mole
Ca/6.023*10^23 atoms)(40g Ca/1mol Ca) = <span>4.45*10^13 grams Ca</span>
