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salantis [7]
3 years ago
12

Which electron requires the most ionization energy - the 1st, 2nd or 3rd electron on a sodium atom? Support your answer using at

omic structure. (EX: energy levels, nuclear charge, attraction).
Chemistry
1 answer:
kicyunya [14]3 years ago
6 0

Answer:

3rd

Explanation:

I dont know how to to this explanation to be honest

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HELP ME, MARKED AS BRAINEST
ELEN [110]

Answer:

The answer is B

Explanation:

2Na + 2H2O = 2NaOH + H2

5 0
3 years ago
True or False: When you measure something with a ruler, you should NOT estimate an extra digit. You only need to report the numb
sveticcg [70]

Answer:

False

Explanation:

It is important to give exact measurements.

5 0
3 years ago
As pH increases, what happens to the hydrogen ion concentration?
levacccp [35]

Answer:

According to libretexts the answer would be B. decreases.

Explanation:

If the hydrogen concentration increases, the pH decreases, causing the solution to become more acidic. This happens when an acid is introduced. ... If the hydrogen concentration decreases, the pH increases, resulting in a solution that is less acidic and more basic

6 0
3 years ago
Read 2 more answers
Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c
uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

4x^{2} = 81.9x^{2} -57.8x +9.66

77.9x^{2} -57.8x+9.66 = 0

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0
3 years ago
Two atoms with the same number of protons but a different number of neutrons are
Luda [366]
They are called isotopes.

Example of isotopes are Hydrogen and deuterium.
Hydrogen is 1 proton and 0 neutrons.
Deuterium is 1 proton and 1 neutron
6 0
4 years ago
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