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2 years ago

Which of the following atoms releases more energy than oxygen when an electron is added to its valence energy level?

1 answer:

5 0

The answer is b. fluorine (F). This is because fluorine is a more electronegative (electron loving) atom compared to oxygen. It is found on the 7th group of the periodic table, so it has 7 valence electrons. It only needs one more electron to achieve the ideal number of valence electrons which is 8. Oxygen, on the other hand, is in group 6 so it needs 2 more electrons to complete its valency. Helium can't be the answer because it is already in group 8, meaning it is already stable and won't release energy.

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R is the midpoint of FG.<br><br> FR = 26 find RG

vichka [17]

Answer:

Explanation:

This is because if R is the midpoint of FRG, FR is half of FRG, so basically all you do it multiply by 2 to get the FRG

7 0

2 years ago

Define the term pH in acid and base

Kaylis [27]

Answer:

a measure of how acidic/base water is.

Explanation:

hope it's helpful (◕ᴗ◕✿)

3 0

1 year ago

Consider dissolving sugar in water and magnesium sulfate in water. is any new substance being produced? what happens after sodiu

Olenka [21]

Answer:

Here's what I get

Explanation:

1. Sugar

(a) Dissolving in water

The white solid dissolves in water to give a colourless solution. There is no evidence that a new substance is being produced.

(b) Addition of sodium hydroxide

Adding the colourless solution of sodium hydroxide to the colourless sugar solution gives a colourless solution. There is no evidence that a new substance is being produced.

2. Magnesium sulfate

(a) Dissolving in water

The colourless crystals dissolve in water to give a colourless solution. There is no evidence that a new substance is being produced.

(b) Addition of sodium hydroxide

Adding the colourless solution of sodium hydroxide to the colourless solution of magnesium sulfate gives a white precipitate (see image). This is evidence that a new substance is being produced.

8 0

2 years ago

What is the significance of the Roman Numeral when naming binary ionic compounds?

serious [3.7K]

Answer:

Roman numbers are oxidation state of the metals

Because some elements of metals show more than one oxidation state like iron Fe2+ in ferrous and Fe3+ in ferric.

Explanation:

8 0

2 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

2 years ago