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Fantom [35]
2 years ago
15

What mass in kilograms of E85( d = 0.758 g/mL ) can be contained in a 13.0 gal tank?Express your answer to three significant fig

ures and include the appropriate units.
Chemistry
1 answer:
melamori03 [73]2 years ago
4 0

The mass in kilograms of E85 with density of 0.758 g/mL that can be contained in a 13.0 gallon tank is 2,869.34g.

<h3>How to calculate mass?</h3>

The mass of a substance can be calculated using the following formula:

Density = mass ÷ volume

According to this question, a substance E85 has a density of 0.758g/mlL and it contains 13.0gallon of liquid. The mass can be calculated as follows:

13 gallon can be converted to 3785.412 millilitres

mass = 0.758 × 3785.412

mass = 2,869.34g

Therefore, the mass in kilograms of E85 with density of 0.758 g/mL that can be contained in a 13.0 gallon tank is 2,869.34g.

Learn more about mass at: brainly.com/question/19694949

#SPJ1

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When multiplying numbers in scientific notation what do you do with exponents?
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4 years ago
Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi
sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Entropy change of the reaction ; ΔS

\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]

=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

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3 years ago
3) Which of the following is a true statement
Hitman42 [59]

Answer:

i need help!!!!!!

Explanation:

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4 years ago
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