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3 years ago

Match each of the following descriptions to thecorrect element listed

Chemistry

1 answer:

Slav-nsk [51]3 years ago

4 0

Answer: Please see explanation for answer

Explanation:Matching the descriptions to the correct element listed gives

Element Group Name Group No Metal/ Nonmetal/ Metalloid Description

Neon Noble gases 18 Non metal Colorless, tasteless, and odorless gas, unreactive, nonconductive.

Bromine Halogens 17 Non metal Not found as a free element (uncombined) in nature, reddish-brown liquid that vaporizes readily at room temperature to a red gas with a strong disagreeable odor.

Beryllium Alklaline earth metal 2 Metal Silvery-white, ductile, malleable, conductive solid with a high melting point for this type of element

Platinum Platimum group metals 10 Metal Silvery white, relatively soft, low density, conductive solid that is not found as a free element in nature but commonly found combined in alloys with copper or nickel

Potassium Alkali Metals 1 Metal Soft, easily cut with a knife to expose a silvery surface that rapidly oxidizes in air; never found uncombined in nature

Silicon Carbon family 1`4 Metalloid Metallic luster and grayish solid, very common in rocks and gemstones such as amethyst and opal, semiconductor

Sulfur Calcogens 16 Non metal Pale yellow, odorless, brittle solid at room temperature, nonconductor

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if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

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

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

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