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3 years ago

Which statement(s) is/are TRUE about covalent bonds?

Chemistry

1 answer:

tresset_1 [31]3 years ago

3 0

Answer:

1. Covalent bonds can form between two nonmetal atoms.

2. Covalent bonds can form between atoms of the same element.

3. Covalent bonds can form between atoms of different elements.

Explanation:

I hope this helps u! :D

Explanation:

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What number is considered neutral? (On the pH scale)

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Enter the Ksp expression for the solid AB2 in terms of the molar solubility x. AB2 has a molar solubility of 3.72×10−4 M. What i

AlekseyPX

Answer:

2.06 × 10⁻¹⁰

Explanation:

Let's consider the solution of a generic compound AB₂.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

We can relate the molar solubility (S) with the solubility product constant (Kps) using an ICE chart.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

I 0 0

C +S +2S

E S 2S

The solubility product constant is:

Kps = [A²⁺] × [B⁻]² = S × (2S)² = 4 × S³ = 4 × (3.72 × 10⁻⁴)³ = 2.06 × 10⁻¹⁰

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How do you express a number in scientific notation?

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Which statement describes a change that occurs during a chemical reaction

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The compound of the chemical is altered

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3 years ago

At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initiall

mestny [16]

Answer:

concentration of $[O_2]$ = 0.0124 = 12.4 ×10⁻³ M

concentration of $[CO]$ = 0.0248 = 2.48 ×10⁻² M

concentration of $[CO_2]$ = 0.4442 M

Explanation:

Equation for the reaction:

$2CO_2_{(g)$ ⇄ $2CO_{(g)$ + $O_2_{(g)$

Concentration of $CO_2_{(g)$ = $\frac{2.3}{4.9}$ = 0.469

For our ICE Table; we have:

$2CO_2_{(g)$ ⇄ $2CO_{(g)$ + $O_2_{(g)$

Initial 0.469 0 0

Change - 2x +2x +x

Equilibrium (0.469-2x) 2x x

K = $\frac{[CO]^2[O]}{[CO_2]^2}$

K = $\frac{[2x]^2[x]}{[0.469-2x]^2}$

$4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}$

Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²

$4.1*10^{-6} = \frac{2x^3}{(0.938)}$

$2x^3 =3.8458*10^{-6$

$x^3 =\frac{3.8458*10^{-6}}{2}$

$x^3=1.9229*10^{-6$

$x=\sqrt[3]{1.9929*10^{-6}}$

x = 0.0124

∴ at equilibrium; concentration of $[O_2]$ = 0.0124 = 12.4 ×10⁻³ M

concentration of $[CO]$ = 2x = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of $[CO_2]$ = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

3 0

3 years ago