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2 years ago

Which of the following conversions is/are incorrect.

Chemistry

1 answer:

Elanso [62]2 years ago

6 0

Answer:

B + C ARE INCORRECT FROM THE ABOVE GIVEN CONVERSIONS.

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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

ira [324]

Explanation:

The given data is as follows.

m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)

Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.

Also, we know that the relation between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

$\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= -554.12 J/K + 647.49 J/K

= 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

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3 years ago

How does matter and energy cycle through an ecosystem?

Ostrovityanka [42]

Answer:

Explanation:

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3 years ago

How to balance lead and silver acetate yields lead (||) acetate and silver

arsen [322]

The given elements put into an equation using their symbols are as follows:
Pb + $AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb + $AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb + $2AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb + $2AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + 2Ag

That is your final equation

The coefficients are 2 + 2 = 1 + 2

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3 years ago

A chemical ________ involves one or more reactants changing into new products.

guapka [62]

Your answer is C. Reaction

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dezoksy [38]

the answer is O reputation hope this helps did this before

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