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GenaCL600 [577]

1 year ago

8

Explain how the models you developed show that when methane combines with oxygen to form carbon dioxide and water, no atoms are

left over from the reaction.

1 answer:

Xelga [282]1 year ago

3 0

Answer: the bonds in the methane and oxygen come apart, the atoms rearrange and then re-bond to form water and carbon dioxide

Explanation:^

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Ancient Egyptians used a variety of lead compounds as white pigments in their cosmetics, including PbS, PbCO3, PbCl(OH), and Pb2

ser-zykov [4K]

Answer : The compound contains the highest percentage of lead (by mass) is, PbS.

Explanation :

To calculate the percentage of lead in sample, we use the equation:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100$

<u>For $PbS$ :</u>

Mass of $PbS$ = 239.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{239.3g}\times 100=86.58\%$

The percentage of lead in the $PbS$ is 86.58 %.

<u>For $PbCO_3$ :</u>

Mass of $PbCO_3$ = 267.2 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{267.2g}\times 100=77.55\%$

The percentage of lead in the $PbCO_3$ is 77.55 %.

<u>For $PbCl(OH)$ :</u>

Mass of $PbCl(OH)$ = 259.7 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{259.7g}\times 100=79.78\%$

The percentage of lead in the $PbCl(OH)$ is 79.78 %.

<u>For $Pb_2Cl_2CO_3$ :</u>

Mass of $Pb_2Cl_2CO3_$ = 545.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{545.3g}\times 100=37.99\%$

The percentage of lead in the $Pb_2Cl_2CO3_$ is 37.99 %.

Hence, from this we conclude that, the compound PbS contains the highest percentage of lead (by mass).

6 0

1 year ago

There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 12.01%. Now that you have the contribution

mestny [16]

Answer: The average atomic mass of X is 16.53

Explanation:

Mass of isotope X-19 = 2.282

% abundance of isotope X-19 = 12.01% = $\frac{12.01}{100}=0.1201$

Mass of isotope X-21 = 18.48

% abundance of isotope X-21 = (100-12.01)% = $\frac{100-12.01}{100}=0.8799$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(2.282\times 0.1201)+(18.48\times 0.8799)]$

$A=16.53$

Therefore, the average atomic mass of X is 16.53

4 0

1 year ago

Na3X, Enter the group number of X?

Katen [24]

X will be in group 5, since if you exchange the valencies of Na with any element on group 5, you will get Na3X

4 0

1 year ago

Read 2 more answers

An angry science student drops his textbook out of a second story window. Why does it fall to the ground? You may

liubo4ka [24]

Answer:

I think qs. "A" the net force acting on the book by gravity is down..

4 0

1 year ago

If the standard free energy change for the conversion of fructopyranose to fructofuranose is 1.7 kJ/mol, what fraction of the to

Dovator [93]

Explanation:

We know that relation between Gibb's free energy and temperature is as follows.

$\Delta G = -RT ln K$

We are given that the value of $\Delta G$ is 1.7 kJ/mol.

And,

k = $\frac{[product]}{[substrate]}$

= $\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}$

Since, k = 0.50357 and temperature is equal to 298 K.

Therefore,

$\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}$ = 0.50357

so, $\frac{\text{[fructofuranose]}}{\text{[fructopyranose]}}$ + 1 = 1.50357

= $\frac{\text{[total fructose solution]}}{\text{[fructopyranose]}}$

= $\frac{1}{1.50357}$

= 0.665

Therefore, we can conclude that 0.665 fraction of the total fructose in solution is fructopyranose.

7 0

1 year ago