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Lubov Fominskaja [6]
3 years ago
7

Give the number of significant figures in this number: 7.8 x 10¹²

Chemistry
1 answer:
denis-greek [22]3 years ago
8 0
The number is written in scientific notation.


The significant figures are the numbers known precisely plus 1 digit that is uncertain.


When the numbers are expressed in scientific notation all the digits before  10^n are significant.


So, 7 and 8 are significant figures and the answer is that the number of significant figures is 2.


Answer: 2
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Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota
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(b)ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = 0 ; total change in entropy = 2.881 J/K

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Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

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Initial volume = V_{1}

Final volume = V_{2} = 2V_{1}

(a) Change in entropy of the system ΔS_{sys} = nRIn\frac{V_{2} }{V_{1} }

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

ΔS_{sys} = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding ΔS_{sur} = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, ΔS_{sys}  = 2.881 J/K

Since surrounding does not change in this process ΔS_{sur} = 0.

total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

ΔS_{sys}  = 0

Since heat energy is not transferred from the system to the surrounding

ΔS_{sur}  = 0

total change in entropy = ΔS_{sys}+ΔS_{sur} = 0

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