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Sergeeva-Olga [200]
3 years ago
11

Please help i will mark brainlist

Chemistry
2 answers:
Sphinxa [80]3 years ago
4 0

Answer:

V= 21 cubic cm (( I am not sure about density ))

Explanation:

V= L×W×H

Elan Coil [88]3 years ago
4 0

Answer:

volume = 21 cubic cm or 21 cm^3

density = 3 g/cm^3

Explanation:

volume = length × breadth × height

  • vol = 7 × 3 × 1 = 21 cubic cm

hence, volume = 21 cubic cm.

density = mass / volume

  • density = 63 / 21 = 3 g / cm^3

<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>. </em><em>.</em><em>.</em><em>.</em>

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How many different elements are combined in sulfur dioxide
Eddi Din [679]

Answer: 2elements sulfur and oxygen

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
A solution containing 292 g of Mg(NO3)2 per liter has a density of 1.108 g/mL. The molality of the solution is:
serg [7]

Answer: D) 2.41 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molality=\frac{n}{W_s}

where,

n = moles of solute

 W_s = weight of solvent in kg

moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{292g}{148g/mol}=1.97moles

volume of solution = 1L = 1000 ml      (1L=1000ml)

Mass of solution={\text {Density of solution}}\times {\text {Volume of solution}}=1.108g/ml\times 1000ml=1108g

mass of solute = 292 g

mass of solvent = mass of solution - mass of solute = (1108- 292) g = 816g = 0.816 kg

Now put all the given values in the formula of molality, we get

Molality=\frac{1.97moles}{0.816kg}=2.41mole/kg

Therefore, the molality of solution will be 2.41 mole/kg

6 0
4 years ago
A certain mass of nitrogen gas is added to a vessel of 405 mL containing 0.52 g oxygen. The
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Answer:

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5 0
3 years ago
When water evaporates
vladimir1956 [14]

c) The motion of water molecules causes them to leave the liquid and become a gas

Explanation:

When water evaporates, the motion of water molecules causes them to leave the liquid and become a gas. Evaporation is merely a phase change in which liquids changes to solid.

  • During the process of evaporation, heat energy causes the motion of the particles or molecules to increase.
  • This will then break the intermolecular forces holding them together leading to more freedom of the molecules.
  • The molecules will then move from being liquids to solids.

Learn more:

Evaporation brainly.com/question/10972073

#learnwithBrainly

5 0
4 years ago
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