All categories

3 years ago

ILL MARK BRAINLIEST :)

Chemistry

1 answer:

DanielleElmas [232]3 years ago

3 0

Explanation:

According to the law of conservation of mass, mass can neither be created nor destroyed but it can simply be transformed from one form to another.

For example, $Na^{+} + Cl^{-} \rightarrow NaCl$

Mass of Na = 23 g/mol

Mass of Cl = 35.5 g/mol

Sum of mass of reactants = mass of Na + mass of Cl

= 23 + 35.5 g/mol

= 58.5 g/mol

Mass of product formed is as follows.

Mass of NaCl = mass of Na + mass of Cl

= (23 g/mol + 35.5) g/mol

= 58.5 g/mol

As mass reacted is equal to the amount of mass formed. This shows that mass is conserved.

As a result, law of conservation of mass is obeyed.

You might be interested in

Order the various instruments used for measuring water in this experiment (balances, graduated cylinders, beaker and pipette) fr

lara [203]

Answer:

(Most accurate) pippete>graduated cylinder>beaker>balance (Least accurate)

Explanation:

<em>Most accurate. A pipette prived the most accurate method for delivering a known volume of solution, for example, a 10mL transfer pipette has an accuracy of ±0.02mL</em>
A graduated cylinder is specifically used to deliver a known volume, its typical accuracy is ±1%, this means that a 100ml graduated cylinder is accurate to ±1mL.
A beaker is a multipurpose cylindrical glass mainly used to hold liquids. Even though they are graduated, these marks are an estimation, the beaker's accuracy is around 10%.
Least accurate. A balance measures an object's mass, even though water's density is close to 1, a balance is not the ideal equipment to measure volume, its capacity usually goes between 100-200grams and can measure mass to the nearest ±0.01mg to ±1mg.

I hope you find this information useful and interesting! Good luck!

4 0

3 years ago

Determine the volume of a piece of gold with a mass of 318.97g

djyliett [7]

Answer:

318 g / 19.32 g v = 16. your volume is 16 hope this helps

Explanation:

6 0

3 years ago

What is one way to decrease carbon?

katrin2010 [14]

Answer:

Buy local and eat a more diversified diet including less meat and dairy to reduce your carbon emissions resulting from the use of fossil fuel-based fertilizers, pesticides, and gas required to produce and transport of the food you eat. Support clean energy sources.

Explanation: or

Alternatives to drivingWhen possible, walk or ride your bike in order to avoid carbon emissions completely. Carpooling and public transportation drastically reduce CO2 emissions by spreading them out over many riders.

6 0

3 years ago

Which term best describes the role of hydrogen gas in the formation of water molecules?

kobusy [5.1K]

The answer is D, reactant.

6 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

2 years ago