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Dennis_Churaev [7]
2 years ago
10

How many gallons of 80% antifreeze solution must be mixed with 60 gallons of 20% antifreeze to get a mixture that is 70% antifre

eze? Use the six-step method.
...?
Chemistry
2 answers:
Sveta_85 [38]2 years ago
7 0
<span>7X+.25*100=.6(100+X)
.7X+25=60+.6X
.7X-.6X=60-25
.1X=35
X=35/.1

X=350 GALLONS OF 70% ANTIFREEZE WILL BE NEEDED.
PROOF
.7*350+25=.6*450
245+25=270
270=270</span>
Anarel [89]2 years ago
6 0
Let A be the 80% solution and B be the 20% solution and P be the produce solution of 70%. Va and Vb and Vp are the volumes of A and B and P respectively.
Va + 60 = Vp
0.7Vp = 0.8Va + 0.2(60)
Substituting the value of Vp from the first equation:
0.7(Va + 60) = 0.8Va + 12
30 = 0.1Va
Va = 300 gallons
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masha68 [24]

Answer:

\boxed {\boxed {\sf 0.0018 \ mol \ H_2 O }}

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

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There are 2 hydrogen atoms, so multiply the mass by 2.

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Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

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Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

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Multiply. The grams of H₂O will cancel out.

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The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

0.0018 \ mol \ H_2 O

There are about <u>0.0018 moles in 0.032 grams.</u>

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