Hello I need help with this question

A number of valence electrons can be determined by looking at the... *

inessss [21]

Answer:

The main group number for an element can be found from its column on the periodic table.

Explanation:

For example, carbon is in group 4 and has 4 valence electrons. Oxygen is in group 6 and has 6 valence electrons.

8 0

3 years ago

The equilibrium constant, K, for the following reaction is 1.80x102 at 698 K. 2HI(9) =H2(g) +129) An equilibrium mixture of the

Mrrafil [7]

Explanation:

The equilibrium constant of the reaction = $K_c=1.80\times 10^{-2}$

Initial concentration of HI = 0.311 M

After addition 0.174 mol of HI(g)

Concentration of HI added = $\frac{0.174 mol}{1 L}=0.174 M$

New concentration of HI = 0.311 M + 0.174 M = 0.485 M

$2HI\rightleftharpoons H_2+I_2$

Initial concentration:

0.311 M $4.71\times 10^{-2} M$ $4.71\times 10^{-2} M$

At equilibrium:

(0.485 M - x) $(4.71\times 10^{-2} M+x)$ $(4.71\times 10^{-2} M+x)$

$K_{eq}=\frac{[H_2][I_2]}{[HI]^2}$

$1.80\times 10^{-2}=\frac{(4.71\times 10^{-2} M+x)\times (4.71\times 10^{-2} M+x)}{ (0.485 M-x)^2}$

$\sqrt{1.80\times 10^{-2}}=\frac{(4.71\times 10^{-2} M+x)}{ (0.485 M-x)}$

$0.1342=\frac{(4.71\times 10^{-2} M+x)}{(0.485 M-x)}$

$0.065087 M-0.1342x=4.71\times 10^{-2} M+x$

$0.065087 M-4.71\times 10^{-2} M=1.1342x$

$x=\frac{0.017987 M}{1.3142}=0.01586 M$

Equilibrium concentrations:

$[HI]=0.485 M-x = 0.485 M - 0.01586 M= 0.46914 M$

$[H_2]=[I_2]=4.71\times 10^{-2} M+x=4.71\times 10^{-2} M+0.01586 M$

$[H_2]=[I_2]=0.06296 M$

3 0

3 years ago

Hemos preparado la siguiente disolución 40 g de sal en 250 g de agua. Escribe:

Anestetic [448]

Calcula el Tanta por ciento en peso de soluto

5 0

4 years ago

For the following reaction, 42.2 grams of potassium hydrogen sulfate are allowed to react with 21.4 grams of potassium hydroxide

ASHA 777 [7]

Answer:

53.99g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

KHSO4(aq) + KOH(aq) —> K2SO4(aq) + H2O(l)

Step 2:

Determination of the masses of KHSO4 and KOH that reacted and the mass of K2SO4 produced from the balanced equation.

This is illustrated below:

Molar mass of KHSO4 = 39 + 1 + 32 + (16x4) = 136g/mol

Mass of KHSO4 from the balanced equation = 1 x 136 = 136g

Molar mass of KOH = 39 + 16 + 1 = 56g/mol

Mass of KOH from the balanced equation = 1 x 56 = 56g

Molar mass of K2SO4 = (39x2) + 32 + (16x4) = 174g/mol

Mass of K2SO4 from the balanced equation = 1 x 174 = 174g.

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH to produce 174g of K2SO4

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH.

Therefore, 42.2g of KHSO4 will react with = (42.2 x 56)/136 = 17.38g of KOH.

From the above calculations, we can see that only 17.38g out of 21.4g of KOH given was needed to react completely with 42.2g of KHSO4.

Therefore, KHSO4 is the limiting reactant and KOH is the excess reactant.

Step 4:

Determination of the maximum mass of K2SO4 produced from the reaction.

In this case, the limiting reactant will be used as all of it is used up in the reaction. The limiting reactant is KHSO4 and the maximum amount of K2SO4 produced can be obtained as follow:

From the balanced equation above, 136g of KHSO4 reacted to produce 174g of K2SO4.

Therefore, 42.2g of KHSO4 will react to produce = (42.2 x 174)/136 = 53.99g of K2SO4.

Therefore, the maximum amount of K2SO4 produced is 53.99g.

8 0

3 years ago

A firework exploding is an example of a

finlep [7]

The correct answer would be a physical change

4 0

4 years ago