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NISA [10]
4 years ago
7

In a vacuum bottle, 320 g of water and 120 g of ice are initially in equilibrium at 0.00∘∘C. The bottle is not a perfect insulat

or. Over time, its contents come to thermal equilibrium with the outside air at 25.0∘∘C. 1) How much does the entropy of universe increase in the process?
Chemistry
1 answer:
shusha [124]4 years ago
3 0

Answer:

ΔS_{univ}=267.7J/K

Explanation:

Hello,

To find the change in the entropy of the universe, we must take into account the following entropy balance:

ΔS_{univ}=\frac{Q_{water}+Q_{ice}}{T_{univ}}

We can stand the universe as the surroundings so the T_{univ} equals the outside air temperature. Then, we must compute both the water's and ice's released heat due to their interaction with the "hot" air:

Q_{water}=320g*4.18J/g^0C*(25-0)^0C= 33440J

Q_{ice}=m_{ice}*(H_{melt,ice}+Cp_{ice}*(T_{univ}-T_{initial}))\\Q_{ice}=120g*(333.7J/g+2.11J/g^0C*(25-0)^0C)\\Q_{ice}=46374J

Finally, we obtain the change (increasing) in the entropy of the universe as follows:

ΔS_{univ}=\frac{33440J+46374J}{(25+273.15)K}

ΔS_{univ}=267.7J/K

* All the data were extracted from Cengel's thermodynamics book.

Best regards.

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ANTONII [103]

Answer:

4m/s in the direction of the turn

Explanation:

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Assuming the runner stays the same speed as he turns, his velocity will be 4m/s in the speed he turns.

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3 years ago
PLEASE HELP ME URGENT!!! How do i find mass when not given in chemistry
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Answer:

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3 years ago
Consider a solution prepared by mixing the following:50.0 mL of 0.100 M Na3PO4100.0 mL of 0.0500 M KOH200.0 mL of 0.0750 M HCl50
Juliette [100K]

Answer:

you must add 50 mL

Explanation:

Hi

KOH is a strong base and by adding 100mL 0.05M you will have an amount of 5 millimol.

NaCN is a base and by adding 50 mL 0,150 M you will have an amount of 7,5 mmol.

HCl is a acid and by adding 200 mL 0,075 M you will have an amount of 15 mmol.

The acid reacts with the bases leaving 2.5 mmol unreacted.

Na3PO4 is a base and by adding 50 mL 0,1 M you will have an amount of 5 mmol.

The 2.5 mmol of acid react with the base PO4 ^ -3 forming a regulatory solution of PO4 ^ -3 and HPO4 ^ -2 of pKa 2.12

5 mmol of acid (HNO3) must be added to obtain a regulatory solution formed by the same amount of HPO4 ^ -2 (2.5 mmol) and H2PO4 ^ -1 (2.5 mmol) with pKa 7.21

Considering a quantity of 5 mmol of HNO3 of concentration 0.1 M, 50 mL must be added.

To calculate the pH of the regulatory solution you should consider pH = pKa × Ca / Cb pH = 7.21 × 2.5 / 2.5 = 7.21 Being in the same solution the volume is the same and can be simplified to achieve a faster calculation.

successes with your homework

5 0
4 years ago
If 100 ml of some pb(no3)2 solution is mixed with 100 ml of 6.50 x 10−2 m nacl solution, what is the maximum concentration of th
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The concentration of Cl⁻ is:
[Cl⁻] = (100 / 200) * (6.5 x 10⁻² M) = 0.0325 M
So:
Ksp = 2.00 x 10⁻⁵ = [Pb²⁺] [Cl⁻]² = (0.5X) * (0.0325)²
X = (2.00 x 10⁻⁵) / (5.28 x 10⁻⁴) = 0.038 M
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4 years ago
The gametes are identical to each other and combine to make an identical organism.
Sliva [168]

Answer:

true

Explanation:

4 0
3 years ago
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