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3 years ago

Salt dissolves in water because salt is held together by:

1 answer:

mote1985 [20]3 years ago

7 0

Water can dissolve salt because the positive part ofwater molecules attracts the negative chloride ions and the negative part of water molecules attracts the positive sodium ions. The amount of a substance that candissolve in a liquid (at a particular temperature) is called the solubility of the substance.

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Chemistry Stoichiometry Question

Zielflug [23.3K]

The correct answer is 1.1 moles

8 0

2 years ago

What is the iupac name of the following compound? <br><br> 2,2-dimethyl-4-ethylheptane

Lady bird [3.3K]

Answer:

The IUPAC name of the compound has already been given which is 2,2-dimethyl-4-ethylheptane.

Explanation:

The IUPAC (International Union of Pure and Applied Chemistry) is an authority in chemistry that provides a guideline and standardized methods in the naming of compounds formed from the periodic table.

In order the give an IUPAC name to a compound, certain steps needs to be followed, these includes:

--> Identify the functional group in the compound as this will form the suffix. For example if the functional group is an alkane the suffix will be -ane.

--> Identify the longest carbon chain (it may not be a straight chain) that contains the functional group. This forms the prefix. Example: if the longest carbon chain is 7 carbon atoms then the prefix will be hept-

--> All the carbons of the longest chain should be numbered

--> Identify branched groups on the chain and name them according to the number of carbon atoms. They usually end with -yl.

--> Finally, combine the elements of the name is a single word.

The structural formula of the IUPAC compound can be found in the attached file for a better understanding. The branched groups are circled.

4 0

3 years ago

PLSSS HELP FAST

Aliun [14]

Note the formula

$\boxed{\sf Mass\:no=No\:of\:protons+No\:of\:neutrons}$

For X:-

No of protons:-

$\\ \sf\longmapsto 19-10=9protons$

It is Fluorine.

For Y:-

No of protons:-

$\\ \sf\longmapsto 16-8=8protons$

It is Oxygen.

Option D is correct

7 0

2 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

Is the electrolysis of water an oxidation-reduction reaction??

QveST [7]

Answer:

Yes

Explanation:

Electrolysis is an example of redox reaction because reduction takes place at cathode and oxidation takes place at anode and both of these reactions take place simultaneously.

8 0

3 years ago