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3 years ago

Are there Muslims? For Muslims

Chemistry

2 answers:

gladu [14]3 years ago

7 0

Answer:

there are Muslims i am also a Muslim

Explanation:

nydimaria [60]3 years ago

6 0

Answer:

i am muslim! waaleikumassaslam

Explanation:

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Murrr4er [49]

Answer:

Look at the properties of Oxygen and Silicon - the two most abundant elements in the Earth's crust - by clicking on their symbols on the Periodic Table.

Explanation:

7 0

3 years ago

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When a material in the liquid state is vaporized and then condensed to a liquid, the steps in the process are, respectively,?

Ahat [919]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the choices that can be found from other sources.
The answer is C endothermic and exothermic

<span>a. exothermic and exothermic
b. exothermic and endothermic
c. endothermic and exothermic
d. endothermic and endothermic</span>

7 0

3 years ago

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The isotope lead-208 (208P)

USPshnik [31]

It is a stable isotope of lead metal. this isotope is of nonradioactive nature. The isotope occur naturally but can also be produced by fission reaction.

It appears as gray powder. With molecular weight of 207.9g/mol.

The boiling point is 1740° while the melting point is 327.4°

It is immediately available as typical and in custom packing of different grade of food, agriculture and pharmaceuticals.

it is used for biomedical and biological labelling. It is also used to measure pb (lead) level in blood by glycogen isotope dilution technique.

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

What color would the methyl orange indicator turn in a solution with a pH of 2?

alexdok [17]

Answer:

The answer is red.

Explanation:

Methyl orange is an organic dye that used as an indicator in acid-base titrations because its color is changed according to pH of the medium (acidic, neutral or basic).
Methyl orange is red in acidic solutions, orange in neutral solutions, and yellow in basic solutions.

Since pH of the solution is 2, the solution is acidic, the color of the solution will be red.

4 0

3 years ago