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Helga [31]
4 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:
Maslowich4 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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Elden [556K]

Answer:

ΔG° = -5.4 kJ/mol

ΔG = 873.2 J/mol = 0.873 kJ /mol

Explanation:

Step 1: Data given

ΔG (NO2) = 51.84 kJ/mol

ΔG (N2O4)  = 98.28 kJ/mol

Step 2:

ΔG = ΔG° + RT ln Q

⇒with Q = the reaction quatient

⇒with T = the temperature = 298 K

⇒with R = 8.314 J / mol*K

⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2 )

⇒ ΔG° = 98.28 kJ/mol - 2* 51.84  kJ/mol

⇒ ΔG° = -5.4 kJ/mol

Part B

ΔG =  ΔG° =RT ln Q

⇒with G° = -5.4 kj/mol = -5400 j/mol

⇒ with R = 8.314 J/K*mol

⇒with T = 298 K

⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577

ΔG = -5400 + 8.314 * 298 * ln(12.577)

ΔG = -5400 + 8.314 * 298 * 2.532

ΔG = 873.2 J/mol = 0.873 kJ/mol

3 0
3 years ago
Choose the atom with:
nata0808 [166]

Answer:

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b)O

c)F

Explanation:

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b)oxygen ion is larger than Na because o have fewer proton

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valentina_108 [34]

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3 0
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stich3 [128]
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8 0
2 years ago
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Answer:

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Explanation:

I hope this will help you :)

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