Which metal in Period 5 is very reactive and has two valence electrons in each atom?

2 answers:

8 0

The answer is Strontium(Sr). The reactive increase from right to left. And this element has two valence electrons. So Rb is not correct. Then the very reactive metal is Sr.

irakobra [83]3 years ago

8 0

Answer: The correct answer is Strontium.

Explanation:

Reactivity of an element is defined as the tendency to loose or gain electrons. Reactivity decreases as we move from left to right in a period and it increases as we move from up to down in a group.

A metal which belongs to Period 5 and has 2 valance electrons is Strontium. This element belongs to Group 2.

The electronic configuration of strontium (Sr) = $[Kr]5s^2$

Hence, the correct answer is Strontium.

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1.04 ⨯ 10 $^{-4}$ M

<h3>Explanation</h3>

A = ε $\cdot l \cdot c$ by the Beer-Lambert law, where

A the absorbance,
$l$ the path length,
ε the molar absorptivity of the solute, and
$c$ concentration of the solution.

A and ε are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$ . The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.