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tino4ka555 [31]
3 years ago
6

Rosa preparo una disolución de agua salada mezclando 120 mg de agua con 5 mg de sal comun ¿ que concentración tiene la disolució

n que preparo ?
Chemistry
1 answer:
enot [183]3 years ago
3 0

Concentration of the solution (% mass) = 4%

<h3>Further explanation</h3>

<em>Rosa prepared a salt water solution by mixing 120 mg of water with 5 mg of common salt, what concentration is the solution she prepare?</em>

<em />

<em />\tt \%mass=\dfrac{mass~solute}{mass~solution}\times 100\%<em />

mass solute=mass of salt=5 mg

mass solution :

mass solute(salt)+mass solvent(water) :

\tt 5~mg+120~mg=125~mg

\tt \%mass=\dfrac{5~mg}{125~mg}\times 100\%\\\\\%mass=4

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Group 7A (17) elements can gain one electron to attain a noble gas electron configuration. In combination with metal atoms that
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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.

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(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
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