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WITCHER [35]
2 years ago
7

1 pts

Chemistry
2 answers:
Paraphin [41]2 years ago
7 0
I thinkkkk it’s OCHbl
pav-90 [236]2 years ago
3 0
It is OCHBI I can confirm
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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
50.0 g of NaNO3 are dissolved into enough water to make 250 mL of solution. What is the molarity of this solution?
d1i1m1o1n [39]

Answer:

2.35 M

Explanation:

Molarity is mol/L of solution. We have to convert the g to mol and the mL to L. G to mol uses the molar mass of the compound. The molar mass of NaNO₃ is 85.00g/mol.

50.0gNaNO3*\frac{1molNaNO3}{85.00gNaNO3} = 0.588molNaNO3

Then you have to convert mL to L.

250mL*\frac{1L}{1000mL} = 0.250L

Now divide the mol by the L.

\frac{0.588mol NaNO3}{0.250L} = 2.352 M

Round to the smallest number of significant figures = 2.35M

7 0
3 years ago
Can someone please helpppp!
dolphi86 [110]

Answer: A

Aloha is the one that loses a helium

5 0
3 years ago
Should the density of a metal depend on its volume?
jekas [21]

Answer:

Density is the ratio of the mass of an object to its volume. Density is an intensive property, meaning that it does not depend on the amount of material present in the sample. ... Since a cubic centimeter is equal to a milliliter, density units can also be expressed as g/mL.

Explanation:

6 0
3 years ago
Fill in the coefficients that will balance the following reaction:
Alex17521 [72]

Answer:

To balance a reaction, the amount of reactants must be equal to the amount of products, as stated by the Law of Conservation of Matter. It may help you to keep track of  the number of each element in a list as you try to balance. It's not able to be balanced.

3 0
2 years ago
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