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3 years ago

Give an example of a purely physical change and an example of a chemical change

Chemistry

1 answer:

vampirchik [111]3 years ago

4 0

Physical change- crunching up paper
Chemical change- burning paper

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Nitrogen dioxide is a red-brown gas that is responsible for the color of photochemical smog. What is the volume of 1 mol of nitr

Ray Of Light [21]

B ideal gas has a volume of 22.4

5 0

3 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

3 years ago

Esterification is what type of reaction?

lyudmila [28]

Esters and Formation of esters. Esters and water are formed when alcohols react with carboxylic acids. This reaction is called esterification, which is a reversible reaction. ... Since esterification is a reversible reaction, esters can undergo hydrolysis to form corresponding alcohol and organic acid.

8 0

3 years ago

Give the formula for an ionic compound formed from each pair of ions

Vera_Pavlovna [14]

Please, find the complete question in the picture attached.

Answer:

1. Na₂O

2. AlF₃

3. MgO

4. Ca₃P₂

Explanation:

1. Na⁺ and O²⁻

An ionic compound will be formed when two ions of opposite charge attract each other.

The positive ion is called cation and the negative ion is called anion.

Thus, every ionic compound has a cation and an anion electrostatically bonded.

The ions must combine in a proportion that genders a neutral compound: so you must have as many cations as negative charge has one anion and as many anions as positive charge has one cation.

This is, if the cation has charge +x, there will be x anions, and if the anion has charge -y, there will be y cations.

Simbolically:

$Cation:A^{+x}\\ \\ Anion:B^{-y}\\ \\ Compound:A_yB_x$

As you see, the subscripts for each element in the chemical formula are obtained by the exchage of the charges of the ions.

Then, for Na⁺ and O²⁻, the subscript for Na will be 2 and the subscript for O will be 1; and the formula of the ionic compound formed by this pair ot ions is:

Na₂O

2. Al³⁺ and F⁻

The subscript of Al will be 1 (because the F ion has charge 1-, and the subscript of F will be 3 (because the Al ion has charge +3).

Thus the ionic compound formed by this pair of ions is:

AlF₃

3. Mg²⁺ and S²⁻

The charge 2+ from Mg atom wil become the subscript 2 of S atom, and the charge 2- will become the subscript 2 of Mg atom:

That results in the formula: Mg₂S₂

Except for some special compounds, the chemical formula is simplified, dividing by the least common denominator. In this case, that means that the two 2 subscripts are simplified to 1, and the final chemical formula for the ionic compound formed by this pair of ions is:

4. Ca²⁺ and P³⁻

The charge 2+ from Ca will become subscript 2 for P and the charge 3- whill become subscript 3 for Ca.

Hence, the ionic compound formed by these two ions is:

Ca₃P₂

7 0

3 years ago

Most elements are ____________. The _____________ elements on the left of the stair-step line on the ______________ of Elements

Vlada [557]

Answer:

metals, 88 , periodic table , luster , heat, electricity , thin, right, opposite , appearance , luster , malleable , ductile, opposite , shiny, moderately

Explanation:

metals, 88 , periodic table , luster , heat, electricity , thin, right, opposite , appearance , luster , malleable , ductile, opposite , shiny, moderately Is the answer.

(Hope this helps can I pls have brainlist (crown)☺️)

4 0

2 years ago