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3 years ago

Which of these statements about neutrons is true?

1 answer:

galina1969 [7]3 years ago

4 0

The answer is B.
A cannot be the answer because only electrons move around the nucleus.

C cannot be it because electrons have a lot more mass than neutrons.

D can't be it because it 's not a typical occurrence for the amount of neutrons to equal the amount of electrons

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Large sandbars along the coastline that form from deposition of

olga55 [171]

Barrier island refers to the large sandbars along the coastline that form

from deposition of sand over long period of time.

<h3>Features of a Sandy environment </h3>

The features of a sandy environment include the following:

Spits

Barrier islands

Tombolos

Barrier Island are formed when sand is deposited over a long time and this

helps to prevent erosion along the coastline area.

Read more about Barrier Island here brainly.com/question/15750467

3 0

2 years ago

What are metals that have been mixed together called?

Svetach [21]

<span> metals that have been mixed together called alloy</span>

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3 years ago

Label the following changes as CHEMICAL or PHYSICAL change

skelet666 [1.2K]

1. Chemical
2. Physical

1. Exothermic
2. Endothermic

1. Chemical
2. Physical

8 0

4 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago

After decaying for 48 hours, l/16 of the original mass of a radioisotope sample remains unchanged. What is the half-life of this

ivolga24 [154]

After the first half-life, 1/2 is unchanged. After the 2nd one, 1/4 remains. After the 4th one, 1/16 remains. So you have 4 half-lives in 48 hours, the half-life is thus 12 hours.

3 0

3 years ago