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Sveta_85 [38]
2 years ago
12

Sodium (Na) and potassium (K) are in the same group in the periodic table. Sodium is in the 11th position. How many valence elec

trons does potassium have?
0



1



9



11
Chemistry
2 answers:
lapo4ka [179]2 years ago
8 0
It's 1 because there is only one electron on the outer shell.
Studentka2010 [4]2 years ago
5 0

Answer: Option (b) is the correct answer.

Explanation:

As it is known that both sodium and potassium are same group elements. Both of then belongs to group 1A.

Since, sodium is at 11th position in the periodic table and its atomic number is also 11. Therefore, electronic distribution of sodium is 2, 8, 1.

Similarly, potassium is placed at 19th position in the periodic table and its atomic number is 19. So, electronic distribution of potassium is 2, 8, 8, 1.

It is known that valence electrons are the electrons present in the outermost shell of an atom.

So, in a potassium atom there is only one valence electron is present.

Thus, we can conclude that potassium have 1 valence electron.

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Answer:

C. The reaction that forms α-1, 6 linkages is catalyzed by branching enzymes.

Explanation:

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This branching enzyme attaches a string of seven glucose units to the carbon at the C-6 position on the glucose unit, forming the α-1,6-glycosidic bond.

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How to draw Hess' Cycle for this question ?
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Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

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which combination of temperature and pressure correctly describes standard temperature and pressure, STP? A) 0 degrees C and 101
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Answer: A


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