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1 year ago

PLS I WILL GIVE BRAINLY How do the number of violence electrons of the light metals compared to those heavier than boron

Chemistry

1 answer:

SCORPION-xisa [38]1 year ago

4 0

The valence electron of the light metals are all in the range of 1 to 3.

<h3>What are Valence electron ?</h3>

Valence Electron is the number of electrons in the outer shell of an atom which participate in bond formation.

Number of Valence Electron in Boron is 3 and other light metals lighter than Boron have 1 , 2 of Lithium and Beryllium respectively,

The valence electron of the light metals are all in the range of 1 to 3.

To know more about Valence Electron

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An organic chemist measures the temperature T of a solution in a reaction flask. Here is the result. T = 149.206 °C Convert T to

jolli1 [7]

<u>Answer:</u> The temperature of the solution in Kelvins is 422.356 K

<u>Explanation:</u>

Temperature is defined as the measure of coldness or hotness of a body. It also determines the average kinetic energy of the particles in a body.

This term is expressed in degree Celsius, degree Fahrenheit and Kelvins. All these units are interchangeable.

The S.I unit of temperature is Kelvins.

We are given:

Temperature of a solution = $149.206^oC$

Conversion used to convert degree Celsius and Kelvins is:

$T(K)=[273.15+T(^oC)]$

$T(K)=273.15+149.206\\T(K)=422.356K$

Hence, the temperature of the solution in Kelvins is 422.356 K

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3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

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IRISSAK [1]

Answer:

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