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9966 [12]
4 years ago
7

Given that oxygen see saying and I was just going to 18 both have an atomic number of 8 how many electrons protons and neutrons

does these oxygen atoms contain
Chemistry
1 answer:
oksian1 [2.3K]4 years ago
5 0
Number of protons = atomic number, assumed number of electrons = number of protons (all elements in the periodic table are assumed to have a neutral charge, in which case protons and electrons are equal in number. If this is not the case they will be ions of a certain element and wil have a + or - sign next to them.), number of neutrons= atomic wt - atomic number.  Oxygen has an atomic number of 8 and an atomic wt of 16 so protons = 8, electrons = 8 and neutrons = 16-8, or 8


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The gain of electrons by an element or ion in a reaction is called __.
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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,

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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.

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