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3 years ago

5

Butane C4 H10 (g),(mc016-1.jpgHf = –125.7), combusts in the presence of oxygen to form CO2 (g) (mc016-2.jpgHf = –393.5 kJ/mol),

and H2 O(g) (mc016-3.jpgHf = –241.82) in the reaction:
a. -5,314.8 kj /mol
b. -2, 657.4 kj / mol
c. 2,657.4 kj / mol
d. 5,314.8 kj / mol

2 answers:

Mice21 [21]3 years ago

4 0

once you do all the math the answer is c

Ira Lisetskai [31]3 years ago

3 0

Answer:

2,657.4 kJ/mol

Explanation:

bc i know all

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g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the

weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

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2 years ago

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AnnyKZ [126]

Answer: Option (4) is the correct answer.

Explanation:

When sulfuric acid reacts with calcium hydroxide then it results in the formation of calcium sulfate and water.

The chemical reaction for the same will be as follows.

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3 years ago

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The heat of fusion AH, of ethyl acetate (C4H802) is 10.5 kinol. Calculate the change in entropy as when 398. g of ethy, acetate

Hitman42 [59]

<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ethyl acetate = 398 g

Molar mass of ethyl acetate = 88.11 g/mol

Putting values in above equation, we get:

$\text{Moles of ethyl acetate}=\frac{398g}{88.11g/mol}=4.52mol$

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=n\times \frac{\Delta H_{fusion}}{T}$

where,

$\Delta S$ = Entropy change = ?

n = moles of ethyl acetate = 4.52 moles

$\Delta H_{fusion}$ = enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the system = $84.0^oC=[84+273]K=357K$

Putting values in above equation, we get:

$\Delta S=\frac{4.52mol\times 10500J/mol}{357K}\\\\\Delta S=132.9J/K$

Hence, the entropy change of the ethyl acetate is 133. J/K

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