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3 years ago

how many electrons is Fluorine likely to receive from the atom of another element during the formation of a bond.

1 answer:

8 0

Answer: Fluorine contains seven valence electrons. It needs one more electron to complete its octet. The oxidation number is 1- and indicates fluorine will gain or share one electron when it combines with another atom to become a more stable compound.

Explanation: Fluorine contains seven valence electrons. It needs one more electron to complete its octet. The oxidation number is 1- and indicates fluorine will gain or share one electron when it combines with another atom to become a more stable compound.

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Which property of water explains why seawater contains over 20 different elements? select one:

Gnesinka [82]

I believe the awnser is a. universal solvent.

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3 years ago

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How many mL of a 1.48 M calcium hydroxide solution are needed to neutralize 36.0 mL of a 1.63 M hydrochloric acid solution

Lelechka [254]

The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL

<h3>Balanced equation </h3>

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 2
The mole ratio of the base, Ca(OH)₂ (nB) = 1

<h3>How to determine the volume of Ca(OH)₂ </h3>

Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
Volume of acid, HCl (Va) = 36 mL
Molarity of acid, HCl (Ma) = 1.63 M
Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(1.63 × 36) / (1.48 × Vb) = 2

58.68 / (1.48 × Vb) = 2

Cross multiply

2 × 1.48 × Vb = 58.68

2.96 × Vb = 58.68

Divide both side by 2.96

Vb = 58.68 / 2.96

Vb = 19.8 mL

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2 years ago

The pictures below show water in a glass and oil in a beaker.

Archy [21]

Answer: is letter D. Because minerals are solid.

Explanation:

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4 years ago

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Which of the following materials is useful for making molds because it has a low melting point?

inessss [21]

<span>Molds are created to achieve a specific design of a material. These materials either came from a process of having a higher or lower temperature. Therefore, the molder must have thermal resistant properties. Low melting points means that the material to be shaped came from a cooler process. Wood and metal have higher thermal conductivity and therefore can easily be cooled. The wax can turn really hard and can be unbreakable when present in colder materials due to the lipids present in it. Clay however can become a mold because of its low melting point.</span>

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3 years ago

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A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .