All categories

2 years ago

Question 2 of 30

Chemistry

1 answer:

icang [17]2 years ago

7 0

Answer: D

Explanation:

Reduction involves the gaining of electrons.

This eliminates choices A and C.

However, charge must be balanced, and this is not the case for choice B, but it is the case for choice D

You might be interested in

How a mineral's chemical make-up looks like?

sveticcg [70]

You would have to go to sephora and check it out

5 0

4 years ago

Explain why hydrogen can only have 2 valence electrons around it when it bonds to other items. What is the maximum number of bon

lys-0071 [83]

Answer:

In first shell only 2 electrons are present in hydrogen only one electron is present in valence shell. To complete its duplet hydrogen can share lose or gain only one electron to form chemical bond.

Hydrogen can form only one bond.

Explanation:

7 0

3 years ago

An atom that contains 22 protons,

Westkost [7]

Answer: 45

Explanation:

Recall that mass number = number of protons + number of neutrons. Hence, the mass number of the atom

= 22 + 23

= 45

Furthermore, the atomic number is equal to the number of protons. Hence, since the number of protons is 22. the atom has an atomic number of 22

Thus, An atom that contains 22 protons,

23 neutrons, and 22 electrons would have a mass of approximately 45

6 0

3 years ago

Read 2 more answers

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

Points

ExtremeBDS [4]

The answer is D. A compound

6 0

3 years ago