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3 years ago

1. Take a look at all the elements in period 3. What do you notice? *

Chemistry

2 answers:

Vedmedyk [2.9K]3 years ago

5 0

Answer:

Explanation:

beacause all the elements in period 3 has different valence electrons hence you look at the first element in period 3 is sodium which has a valence electron of 1 when you go across the period to the right their valence electron decreases to be negative

Lelu [443]3 years ago

4 0

Answer:

D is the correct answer i hope this is correct

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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

Convert 72g of O2 to moles

Volgvan

Moles = mass / molar mass
molar mass of O2 is 32
therefore moles = 72/32
= 2.25 moles

4 0

3 years ago

Read 2 more answers

What is the difference between an enthalpy driven reaction and entropy driven reaction?

Crazy boy [7]

<span>Enthalpy is regarding the amount of heat that is given off or taken in during the process of a reaction, while entropy is about the disorderliness of a reaction.
Both are related in the equation ∆G=∆H-T∆S, where ∆G is the Gibbs free energy. So we can say that a reaction is both enthalpy and entropy driven. It's like, both of them are interlinked with each other. </span>

5 0

3 years ago

I need answer for this pls

Bad White [126]

Answer:

cell membrane

Explanation:

6 0

2 years ago

Read 2 more answers

How much 2.0 M NH4NO3 is needed to make 0.585 L of 1.2 M NH4NO3 solution?

Nutka1998 [239]

Answer:

b . 0.351 L.

Explanation:

Hello!

In this case, since diluted solutions are prepared by adding an extra amount of diluent to a stock-concentrated solution, we infer that the number of moles of solute remains the same, therefore we can write:

$C_1V_1=C_2V_2$

Thus, solving for the volume of the stock solution, V1, we obtain:

$V_1=\frac{C_2V_2}{C_1}$

Now, by plugging in the given data we obtain:

$V_1=\frac{1.2M*0.585L}{2.0M}\\\\V_1=0.351L$

Therefore, the answer is b . 0.351 L.

Best regards!

5 0

3 years ago