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3 years ago

Which element has a complete valence electron shell?

2 answers:

8 0

The element that has a complete valence electron shell is argon (Ar) since it is a noble gas in the last group on the periodic table.

sertanlavr [38]3 years ago

8 0

The answer is argon (Ar)

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Use the equation:

Serga [27]

Answer:

polymerization reaction

4 0

3 years ago

In a solution, the substance that changes phase is the _____, while the substance that does not change phase is the____.

iren [92.7K]

Phase change is the solute
the one that doesn't phase change is the solvent.
A

6 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!!!!

dsp73

Answer: 1. $2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

2. 3 moles of $CuCl_2$ : 2 moles of $Al$

3. 0.33 moles of $CuCl_2$ : 0.92 moles of $Al$

4. $CuCl_2$ is the limiting reagent and $Al$ is the excess reagent.

5. Theoretical yield of $AlCl_3$ is 29.3 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles$

$\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles$

The balanced chemical equation is:

$2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

According to stoichiometry :

3 moles of $CuCl_2$ require = 2 moles of $Al$

Thus 0.33 moles of $CuCl_2$ will require= $\frac{2}{3}\times 0.33=0.22moles$ of $Al$

Thus $CuCl_2$ is the limiting reagent as it limits the formation of product and $Al$ is the excess reagent.

As 3 moles of $CuCl_2$ give = 2 moles of $AlCl_3$

Thus 0.33 moles of $CuCl_2$ give = $\frac{2}{3}\times 0.33=0.22moles$ of $AlCl_3$

Theoretical yield of $AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3$

Thus 29.3 g of aluminium chloride is formed.

5 0

3 years ago

A balloon contains 0.15 moles of co2 gas. how many molecules of co2 are in the balloon?

Sauron [17]

1- One mole is = 6.02 x 10^23 of anything, So one mole of atoms is 6.02x10^23.

2- when the balloon contains 0.15 moles of Co2 gas so:

the no.of molecules of Co2 = 0.15 x 6.02x 10^23

= 9.0 x 10^22

5 0

3 years ago

Read 2 more answers

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago