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kramer
3 years ago
13

What is the percent by mass of phosphorus?

Chemistry
1 answer:
topjm [15]3 years ago
3 0
In what compund is it a part of?

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The experimental apparatus represented above is used to demonstrate the rates at which gases diffuse. When the cotton balls are
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2 years ago
Calculate the pH of a buffer prepared by mixing 20.0 mL of 0.10 M acetic acid and 55.0 mL of 0.10 M sodium acetate
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Answer:

Calculate the pH of a buffer prepared by mixing 30.0 mL of 0.10 M acetic acid and 40.0 mL of 0.10 M sodium acetate.

3 0
3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
You can supply activation energy to begin a reaction by _____.
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If I remember correctly, you would have to heat the reaction beaker over a burner..

I apologize if I'm wrong
3 0
3 years ago
Read 2 more answers
Click on the molecules that represent transpiration.
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Transpiration is the progression of <em>water </em>inside a plant! So, the molecule representing transpiration is going to be good ol' H2O! =)
8 0
3 years ago
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