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Paul [167]
3 years ago
5

Three useful applications of alkaline earth metals

Chemistry
1 answer:
svetlana [45]3 years ago
7 0
Alkaline earth metals refer to a group of elements in the periodic table. They include beryllium, magnesium, calcium, strontium, barium, and radium. Main uses of these elements are given below. Beryllium: It is used in small amounts to make copper beryllium alloy that are very strong and hard.
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All of the alkali earth metals, Group 2, have two valence electrons. Which of these would represent the oxidation number of the
chubhunter [2.5K]
Since Group 2 alkali earth metals have 2 valence electrons, they tend to lose those 2 when forming ionic bonds. And the Loss of Electrons = Oxidation (L.E.O. for short). Therefore this group, including Mg and Ca, have an oxidation of [+2].
So the correct answer is C) +2
5 0
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An area has experienced weathering and erosion for many years. Over time, some rock formations have been left behind forming pea
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Answer:

D.

Explanation:

The rock that is left behind is there because it has resisted the forces of erosion.

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Help thank you so so much!
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In your own words, describe two conditions that are necessary for an eclipse to occur
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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
bazaltina [42]

<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

6 0
3 years ago
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