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SIZIF [17.4K]
3 years ago
10

A radiator contains 10 quarts of fluid, 30% of which is antifreeze. how much fluid should be drained and replaced with pure anti

freeze so that the new mixture is 40% antifreeze?
Chemistry
2 answers:
Svetllana [295]3 years ago
4 0

Answer:C on edge:)

Explanation:

amm18123 years ago
3 0
Let x be the volume of fluid removed and the volume of pure antifreeze that is added. The concentration of antifreeze in the fluid is 0.3, the concentration in pure antifreeze is 1 and that in the final solution is 0.4 The volume of the final solution is 10.
(10 - x)(0.3) + x = 10(0.4)
0.3 + 0.7x = 0.4
x = 1/7 quarts

The volume that should be drained is 1/7 quarts
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Determine the [H3O+] in a 0.265 M HClO solution. The Ka of HClO is 2.9 × 10-8.
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8.8 × 10-5 M is the  [H3O+] concentration in 0.265 M HClO solution.

Explanation:

HClO is a weak acid and does not completely dissociate in water as ions.

the equation of dissociation can be written and ice table to be formed.

 HClO +H2O ⇒ ClO- + H3O+

I  0.265                0        0

C  -x                    +x     +x

E  0.265-x          +x      +x

Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.

Ka = \frac{[ClO-][H3O+]}{[HClO]}

2.9 × 10^-8 = \frac{[x] [x]}{[0.265-x]}

x^{2} = 7.698 x10^{-9}

x = 8.8 × 10-5 M

The hydronium ion concentration is 8.8 × 10-5 M  in 0.265 M solution of HClO.

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2 years ago
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Dafna1 [17]

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6 moles of electrons

Explanation:

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Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

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Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

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