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3 years ago

What oxidation state(s) can the alkaline earth metals exhibit?

Chemistry

1 answer:

iragen [17]3 years ago

6 0

Answer:

Explanation:

Alkaline earth metals are present in group 2 of periodic table. There are six elements in second group. Beryllium, magnesium, calcium, strontium, barium and radium.

All have two valance electrons.

Electronic configuration of Beryllium:

Be = [He] 2s²

Electronic configuration of magnesium.

Mg = [Ne] 3s²

Electronic configuration of calcium.

Ca = [Ar] 4s²

Electronic configuration of strontium.

Sr = [Kr] 5s²

Electronic configuration of barium.

Ba = [Xe] 6s²

Electronic configuration of radium.

Ra = [ Rn] 7s²

They are present in group two and have same number of valance electrons (two valance electrons) and show oxidation state +2 by loosing two valance electrons. They also show similar reactivity.

They react with oxygen and form oxide.

2Ba + O₂ → 2BaO

2Mg + O₂ → 2MgO

2Ca + O₂ → 2CaO

this oxide form hydroxide when react with water,

BaO + H₂O → Ba(OH)₂

MgO + H₂O → Mg(OH)₂

CaO + H₂O → Ca(OH)₂

With sulfur,

Mg + S → MgS

Ca + S → CaS

Ba + S → BaS

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3 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

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<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

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