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Helga [31]
3 years ago
15

If 5.1 l of antifreeze solution (specific gravity = 0.80) is added to 3.8 l of water to make a 8.9 l mixture, what is the specif

ic gravity of the mixture?
Chemistry
1 answer:
Maslowich3 years ago
3 0
Density of the mixture = mass of the mixture / volume of the mixture

Mass of the mixture = mass of antifreeze solution + mass of water.

Mass of antifreeze solution = density of the antifreeze solution * volume

Mass of antifreeze solution = 0.8g/ml * 5.1 l * 1000 ml / l = 4,080 g

Mass of water = density of water * volume of water = 1.0 g/ml * 3.8 l * 1000 ml / l = 3,800 g

Mass of mixture = 4080 g + 3800 g= 7880 g

Volume of mixture = volume of antifreeze solution + volume of water

Volume of mixture = 5100 ml + 3800 ml = 8900 ml

Density of mixture = 7800 g / 8900 ml = 0.876 g/ml

Specific gravity of the mixture = density of the mixture / density of water = 0.876

Answer: 0.876
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A. The patch's area in square kilometers (km²) is 1.61×10⁻⁹ km²

B. The cost of the patch to the nearest cent is 734 cents

<h3>A. How to convert 16.1 cm² to square kilometers (km²)</h3>

We can convert 16.1 cm² to km² as illustrated below:

Conversion scale

1 cm² = 1×10⁻¹⁰ km²

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16.1 cm² = 16.1 × 1×10⁻¹⁰

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brainly.com/question/2139943

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