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4 years ago

Robert uses a spring scale to measure force. What are the units of his measurements? A. Pascals B. Volts C. Joules or D. Newtons

Chemistry

1 answer:

maksim [4K]4 years ago

4 0

Easy it is d for your answer

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Identify the type of interactions involved in each of the following processes taking place during the dissolution of sodium chlo

icang [17]

Answer:

A. Interactions between the ions of sodium chloride (solute-solute interactions).

B. Interactions involving dipole-dipole attractions (solvent-solvent interactions).

C. Interactions formed during hydration (solute-solvent interactions).

D. Interactions involving ion-ion attractions (solute-solute interactions).

E. Interactions associated with an exothermic process during the dissolution of sodium chloride (solute-solvent interactions).

F. Interactions between the water molecules (solvent-solvent interactions).

G. Interactions formed between the sodium ions and the oxygen atoms of water molecules (solute-solvent interactions).

Explanation:

The solution process takes place in three distinct steps:

Step 1 is the separation of solvent molecules.

Step 2 entails the separation of solute molecules.

These steps require energy input to break attractive intermolecular forces; therefore, they are endothermic.

Step 3 refers to the mixing of solvent and solute molecules. This process can be exothermic or endothermic.

If the solute-solvent attraction is stronger than the solvent-solvent attraction and solute-solute attraction, the solution process is favorable, or exothermic (ΔHsoln < 0). If the solute-solvent interaction is weaker than the solvent-solvent and solute-solute interactions, then the solution process is endothermic (ΔHsoln > 0).

In the dissolution of sodium chloride, this process is exothermic.

3 0

3 years ago

C₂H₂ + O₂ CO2 + H2O

Zepler [3.9K]

Answer:

2C2H6 + 7O2 → 4CO2 + 6H2O

2 C2H6O2 + 5 O2 = 4 CO2 + 6 H2O

7 0

3 years ago

Help pls it would be really helpful

Basile [38]

18.is D
19.is A
hope it's correct

3 0

3 years ago

Suppose a star is undergoing a nuclear reaction. What happens to the elements around the star as it begins to emit more and more

AlladinOne [14]

I think the correct answer would be the third option. As the elements around the star begins to emit more and more electromagnetic radiation, the rocky materials are pulled in by the electromagnetic radiation. They are being drawn closer to the star and there would be a very high chance of a nuclear fusion could happen. The pressure and the temperature in a star is so high that it could allow nuclear fusion to happen. As a matter of fact, most of the life of a start is made from hydrogen nuclei fusing together forming a helium nuclei. As it runs out of the hydrogen nuclei, it would fuse other nuclei forming other elements.

3 0

4 years ago

Read 2 more answers

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