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3 years ago

12

Write down the group numbers for carbon,oxygen,nitrogen and sulphur

1 answer:

Bess [88]3 years ago

5 0

Answer:

Carbon - 14

Oxygen - 16

Nitrogen - 15

Sulphur - 16

Explanation:

The question above is related to the "Periodic Table of Elements" which shows the proper arrangement of elements in a table. Every element falls on a<em> group/family</em> within the table. Each group has its own number, and the table has a total of<u> 18 groups</u><em> (from left to right). </em>They are classified according to <em>similarities in their characteristics</em>. For example, group 1 is composed of <em>alkali metals</em> while group 2 is composed of<em> alkali earth metals</em>.

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A sample can not be broken down physically or chemically what is it

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Explanation:

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A target is shielded from a radioactive sample by a sheet of metal. What kind of radiation can pass through the sheet of metal a

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There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc

zhuklara [117]

Answer:

$K_{sp}$ of $Zn(CH_{3}COO)_{2}$ is 0.0513

Explanation:

Solubility equilibrium of $Zn(CH_{3}COO)_{2}$ :

$Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}$

Solubility product of $Zn(CH_{3}COO)_{2}$ ( $K_{sp}$ ) is written as- $K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}$

Where $[Zn^{2+}]$ and $[CH_{3}COO^{-}]$ represents equilibrium concentration (in molarity) of $Zn^{2+}$ and $CH_{3}COO^{-}$ respectively.

Molar mass of $Zn(CH_{3}COO)_{2}$ = 183.48 g/mol

So, solubility of $Zn(CH_{3}COO)_{2}$ = $\frac{43.0}{183.48}M$ = 0.234M

1 mol of $Zn(CH_{3}COO)_{2}$ gives 1 mol of $Zn^{2+}$ and 2 moles of $CH_{3}COO^{-}$ upon dissociation.

so, $[Zn^{2+}]$ = 0.234 M and $[CH_{3}COO^{-}]$ = $(2\times 0.234)M=0.468M$

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3 years ago

Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su

Free_Kalibri [48]

Answer:

Volume = 746 L

Explanation:

Given that:- Mass of copper(II) fluoride = 175 g

Molar mass of copper(II) fluoride = 101.543 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{175\ g}{101.543\ g/mol}$

$Moles_{copper(II)\ fluoride}= 1.7234\ mol$

Also,

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

So,,

$Volume =\frac{Moles\ of\ solute}{Molarity}$

Given, Molarity = 0.00231 M

So,

$Volume =\frac{1.7234}{0.00231}\ L$

<u>Volume = 746 L</u>

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3 years ago