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lubasha [3.4K]
3 years ago
15

A health clinic uses a solution of bleach to sterilize petri dishes in which cultures are grown. the sterilization tank contains

90 gal of a solution of 4% ordinary household bleach mixed with pure distilled water. new research indicates that the concentration of bleach should be 8% for complete sterilization. how much of the solution should be drained and replaced with bleach to increase the bleach content to the recommended level?
Chemistry
1 answer:
Dahasolnce [82]3 years ago
8 0
<span>3.75 gallons need to be drained and replaced with bleach. Change the problem to "What amounts of a 100% solution and a 4% solution is needed to make 90 gallons of a 8% solution?" Given that, we'll use the following values. x = amount of 4% solution. 90-x = amount of a 100% solution. The equation to solve then becomes. 0.04 x + (90-x) = 0.08 * 90 0.04 x + 90 - x = 7.2 Add x to both sides 0.04x + 90 = 7.2 + x Subtract 0.04x from both sides 90 = 7.2 + 0.96x Subtract 7.2 from both sides 82.8 = 0.96x Divide both sides by 0.96 86.25 = x So you now know that you need 86.25 gallons of the original 4% solution and (90-86.25) = 3.75 gallons of the bleach to make the desired 90 gallons. So simply drain 3.75 gallons from the tank and replace with bleach.</span>
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A 25.0 mL sample of a saturated C a ( O H ) 2 solution is titrated with 0.029 M H C l , and the equivalence point is reached aft
Gnesinka [82]

Answer:

0.043 M

Explanation:

The reaction that takes place is:

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As 1 mol of H⁺ reacts with 1 mol of OH⁻, in the 25.0 mL of the Ca(OH)₂ sample there are 1.0817 mmoles of OH⁻.

With that in mind we can <u>calculate the hydroxide ion concentration in the original sample solution</u>, using <em>the calculated number of moles and given volume</em>:

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3 0
3 years ago
A. How many moles of copper equal 8.00 × 109 copper atoms?
BARSIC [14]

Answer:

Explanation:

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no of moles in given no of atoms

= 8 x 10⁹ / 6.02 x 10²³

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102 .5 g calcium

= 102 .5 / 40 moles

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c )

no of moles in 5.04 g lead = 5.04 / 207

= 2.4347 x 10⁻² moles

= 2.4347 x 10⁻²x 6.02 x 10²³ no of atoms of lead

= 14.6568 x 10²¹ no of atoms .

d)  

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2.85 mole = 17.157 x 10²³ atoms .

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moles of fluorine gas = 1.08 x 10³ / 6.02 x 10²³

= .1794 x 10⁻²⁰ moles

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f )

no of moles in .584 g of benzene = .584 / 78

= 7.487 x 10⁻³ moles

no of molecules = 6.02 x 10²³ x  7.487 x 10⁻³

= 45.07 x 10²⁰ molecules .

g )

moles of atoms = 5.09 x 10⁹ / 6.02 x 10²³

= .8455 x 10⁻¹⁴ moles

mass in gram = .8455 x 10⁻¹⁴ x 1

=  .8455 x 10⁻¹⁴ g

h )

.45 moles of Ca₃PO₄ = .45 x 6.02 x 10²³ molecules

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no of atoms of Ca = 3 x 2.709 x 10²³

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