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2 years ago

An atom has two valence electrons, but four shells. This element is a member of which group or family?

Chemistry

1 answer:

Vesna [10]2 years ago

8 0

The correct answer is C: the alkaline earth metals!

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How many moles of oxygen gas occupy 67.2 L of volume at STP? A 0.333 B 3.00 C 1.00

posledela

Answer:

I would say b hope this helps

Explanation:

have a good day btw i had this question before:D

7 0

3 years ago

Chemistry student needs of 55g acetone for an experiment. by consulting the crc handbook of chemistry and physics, the student d

sergeinik [125]

Answer:

70.15 cm³

Solution:

Data Given;

Mass = 55 g

Density = 0.784 g.cm⁻³

Required:

Volume = ?

Formula Used:

Density = Mass ÷ Volume

Solving for Volume,

Volume = Mass ÷ Density

Putting values,

Volume = 55 g ÷ 0.784 g.cm⁻³

Volume = 70.15 cm³

5 0

3 years ago

Why is it important to do further investigations to determine if a chemical change really occurred?

KatRina [158]

Answer:

Explanation:

5 0

3 years ago

A sample of (NH4)2SO4 contains 0.750 mole. what is the mass of the sample

steposvetlana [31]

The answer is 99.10.

5 0

2 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

7 0

3 years ago