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2 years ago

Based on the periodic table why are be, bg, ca and sr in the same column/group/family?

Chemistry

1 answer:

Lesechka [4]2 years ago

7 0

Answer:

The elements are in the same column/group IIA.

See the explanation below, please.

Explanation:

The elements Calcium, Strontium, Beryllium, Magnesium, Barium and Radio, belong to the group of alkaline earth metals located in group IIA of the periodic table, they require 2 electrons to complete their octet (they have 2 valence electrons). reagents than alkali metals.

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Nuclear power plants use a fission reaction of uranium to generate heat to power a generator. A fission reaction is a chain reac

LekaFEV [45]

ANSWER
B.

Explanation : it’s B because the control rods used are usually made of boron which boron is a non fissile material hence won’t undergo fission when bombarded or collided with neutrons.

7 0

1 year ago

Metals (M) in group 2 of the periodic table form ionic bonds

egoroff_w [7]

Answer:

Explanation:

Group two elements are alkaline earth metal.

All these have two valance electrons. In order to achieve noble gas configuration it loses its two valance and get complete octet.

Reaction with group 16.

Reaction with oxygen,

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

5 0

2 years ago

The half-life for the radioactive decay of calcium-47 is 4.5 days. how many half-lives have elapsed after 27 days?

Evgesh-ka [11]

1 half-life - 4.5 days
x half-lives - 27 days

x=(1*27)/4.5=6
6 <span>half-lives have elapsed after 27 days.</span>

8 0

3 years ago

Read 2 more answers

Indicates the kind, number, arrangement, and bonds but not the unshared pairs of the atoms in a molecule.

Vikki [24]

Electronic configuration

3 0

2 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

2 years ago