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Elodia [21]
3 years ago
13

Oo help it’s for a grade

Chemistry
1 answer:
EastWind [94]3 years ago
8 0

Answer: D.kinectic energy

Explanation

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I need help with this question I don’t understand it can someone help me? I’m lost
nirvana33 [79]

Answer:

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Explanation:

5 0
3 years ago
Use the fact that to determine how much the pressure must change in order to lower the boiling point of water by a small amount
erma4kov [3.2K]

Complete Question

Use the fact that d\mu=\frac{V}{N}dp-\frac{s}{N}dT to determine how much the pressure must change in order to lower the boiling point of water by a small amount 3.20e-01 K. You may assume that the entropy and density of the liquid and gas are roughly constant for these small changes. You may also assume that the volume per molecule of liquid water is approximately zero compared to that of water vapor, and that water vapor is an ideal gas. Useful constants: Atmospheric pressure is 101300 Pa The boiling point of water at atmospheric pressure is 373.15 K The entropy difference between liquid and gas per kilogram is 6.05e 03 J/kgK The molecular weight of water is 0.018 kg/mol. (a) 0.00e 00 Pa (b) 1.14e 03 Pa (c) 6.85e 26 Pa (d) 4.24e 05 Pa (e) 3.81e 28 Pa

Answer:

Correct option is B

Explanation:

From the question we are told that:

Given Equation d\mu=\frac{V}{N}dp-\frac{s}{N}dT

Change of boiling point \triangle H=3.20e-01 K

Generally the equation for Change in time is mathematically given by

  d\mu=\frac{V}{N}dp-\frac{s}{N}dT

  dp=\frac{s}{v}dT

Where

    s=Entropy\ difference *molar\ weight

    s=6.05*10^3*0.018j/mol.k

And

    V=\frac{RT}{P} (from ideal gas equation)

Therefore

 dp=\frac{Ps}{RT}dT

 dp=\frac{101300*6.05*10^3*0.018}{8.314*373.15}3.20*10^{-1}

 dp=1137.873pa

 dp=1.14e 03 Pa

Therefore correct option is B

4 0
3 years ago
Need help what is the answer
WARRIOR [948]

Answer:

11.009306 or 11.009

Explanation:

boron 11 has atomic mass of 11.009306

6 0
3 years ago
What is the periodic element symbol for gold?
Makovka662 [10]

Gold is Au on the periodic table

7 0
4 years ago
When 32.5g of carbon is heated with silicone dioxide, 28.8 g of carbon monoxide is produced. What is the percent yield of carbon
Anvisha [2.4K]

Answer:

Percent yield of carbon monoxide is approximately 38%.

Explanation:

Given reaction : C + SiO₂ → Si + CO

On balancing it : 2C + SiO₂ → Si + 2CO

⇒ 2 mole of carbon produces 2 mole of carbon monoxide

or 2×12=24 g of carbon produces 2×(12+16)=56 g of carbon monoxide

⇒ Expected yield of carbon monoxide when 32.5g of carbon is heated with silicon dioxide is 32.5×\frac{56}{24}

⇒ Expected yield = \frac{455}{6} = 75.8 g

Actual yield is given as 28.8 g

⇒ The percent yield of carbon monoxide = \frac{Actual\ yield}{Expected\ yield}×100

= \frac{28.8}{75.8}×100

≅<u>38%</u>

5 0
4 years ago
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