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3 years ago

9

How many valence electrons are in each atom?

1 answer:

Zarrin [17]3 years ago

7 0

Group 1 atoms only have one. Group 2 has 2 valence electrons, and so on.

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What components of an endothermic reaction have a higher heat content

xxTIMURxx [149]

Endothermic reaction is the reaction in which the system absorbs energy. The system absorbs heat and cools the surroundings. Reactants are the substances that are involved in the reaction and at the end of the reaction product are formed.
In the endothermic reaction because heat is absorbed , the products have a higher heat.

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3 years ago

The sample of 15.0 g of KCl is dissolved into a solution with a total volume of 250.0 mL. What is the molarity of KCl in the sol

lina2011 [118]

Answer:

0.805 M.

Explanation:

Hello!

In this case, since the molarity of a solution is computing by dividing the moles of solute over the volume of solution in liters (M=n/V), for 15.0 g of potassium chloride (74.55 g/mol) we compute the corresponding moles:

$n=15.0gKCl*\frac{1molKCl}{74.55gKCl}=0.201molKCl$

Next, since the volume is 0.2500 in liters, the molarity turns out:

$M=\frac{0.201mol}{0.2500L} \\\\M=0.805M$

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3 years ago

What are the 3 principles of the Kinetic molecular theory?

Sever21 [200]

Answer:

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Explanation:answer

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3 years ago

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2 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago