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2 years ago

Write the full electron configuration for the element that is n=4 and has 5 valence electrons.

Chemistry

1 answer:

alukav5142 [94]2 years ago

4 0

Answer:

Greg rusedski is not a fan on his side but answer to village in the future will have to go to a job 6y6666666 and the other two 5 to be 56th 6out and 6 to 56 to 5

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Potassium is isotopic and has RAM of 39.5 work out the percentage abundance of each isotope in a given sample of potassium which

Troyanec [42]

Yo sup??

Let the percentage of K-39 be x

then the percentage of K-40 is 100-(x+0.01)

We know that the net weight should be 39.5. Therefore we can say

(39*x+40*(100-(x+0.01))+38*0.01)/100=39.5

(since we are taking it in percent)

39*x+40*(100-(x+0.01))+38*0.01=3950

39x+4000-40x-0.4+0.38=3950

2x=49.98

x=24.99

=25 (approx)

Therefore K-39 is 25% in nature and K-40 is 75% in nature.

Hope this helps.

3 0

2 years ago

For the equilibrium

Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

$H_2S=0.596M$

$H_2=0.004 M$

$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

Now, equilibrium concentration of gases

$H_2S=0.30+2(0.148)=0.596M$

$H_2=0.30-2(0.148)=0.004 M$

$S_2=0.150-0.148=0.002 M$

3 0

2 years ago

***20 pts***

BartSMP [9]

Answer:

0.131 mol

Explanation:

To convert grams to moles, you need to use the molar mass. The molar mass is the mass number. You can find this on the periodic table. The molar mass of sodium is 22.990 g/mol.

(3.02 g)/(22.990 g/mol) = 0.131 mol