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Molodets [167]
3 years ago
9

What is the amount of ca3(po4)2 that can be prepared from a mixture of 9g of Ca(OH) 2 and 11g of H3po4?

Chemistry
1 answer:
schepotkina [342]3 years ago
6 0
Moles Ca(OH)2 = 12.9 /74.092 g/mol=0.174 
<span>moles H3PO4 = 18.37/98.0 g/mol=0.187 </span>
<span>3 : 2 = x : 0.187 </span>
<span>x = moles Ca(OH)2 needed =0.281 </span>
<span>we have only 0.174 moles of calcium hydroxide so it is the limiting reactant </span>
<span>moles Ca3(PO4)2 = 0.174/3=0.0580 </span>
<span>mass = 0.0580 mol x 310.18 g/mol=18.0 g</span>
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A student has a piece of aluminum metal. Which is the most reasonable assumption the student could make about the metal?
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Read 2 more answers
40POINTS AND BRAINLIEST
mina [271]

Answer:

CI

Explanation:

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7 0
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Be sure to answer all parts. Dimercaprol (HSCH2CHSHCH2OH) was developed during World War I as an antidote to arsenic-based poiso
Sauron [17]

<u>Answer:</u>

<u>For A:</u> The number of arsenic atoms are 3.4\times 10^{21}

<u>For B:</u> The percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

<u>Explanation:</u>

  • <u>For A:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of dimercaprol = 696 mg = 0.696 g    (Conversion factor:  1 g = 1000 mg)

Molar mass of dimercaprol = 124.21 g/mol

Putting values in above equation, we get:

\text{Moles of dimercaprol}=\frac{0.696g}{124.21g/mol}=0.0056mol

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.0056 moles of dimercaprol will contain 0.0056\times 6.022\times 10^{23}=3.4\times 10^{21} number of molecules.

As, 1 molecule of dimercaprol binds with 1 atom of Arsenic

So, 3.4\times 10^{21} number of dimercaprol molecules will bind with = 1\times 3.4\times 10^{21}=3.4\times 10^{21} number of arsenic atoms

Hence, the number of arsenic atoms are 3.4\times 10^{21}

  • <u>For B:</u>

We know that:

Molar mass of dimercaprol = 124.21 g/mol

Molar mass of mercury = 200.59 g/mol

Molar mass of thallium = 204.38 g/mol

Molar mass of chromium = 51.99 g/mol

Also, 1 molecule of dimercaprol binds with 1 metal atom.

To calculate the percentage composition of metal in a complex, we use the equation:

\%\text{ composition of metal}=\frac{\text{Mass of metal}}{\text{Mass of complex}}\times 100     ......(1)

  • <u>For mercury:</u>

Mass of Hg-complex = (200.59 + 124.21) = 324.8 g

Mass of mercury = 200.59 g

Putting values in equation 1, we get:

\%\text{ composition of mercury}=\frac{200.59g}{324.8g}\times 100=61.76\%

  • <u>For thallium:</u>

Mass of Tl-complex = (204.38 + 124.21) = 328.59 g

Mass of thallium = 204.38 g

Putting values in equation 1, we get:

\%\text{ composition of thallium}=\frac{204.38g}{328.59g}\times 100=62.2\%

  • <u>For chromium:</u>

Mass of Cr-complex = (51.99 + 124.21) = 176.2 g

Mass of chromium = 51.99 g

Putting values in equation 1, we get:

\%\text{ composition of chromium}=\frac{51.99g}{176.2g}\times 100=29.51\%

Hence, the percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

8 0
3 years ago
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