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nikklg [1K]
3 years ago
9

Which of these best describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? Which of these bes

t describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? F. Nitrogen monoxide has one more atom of nitrogen. G. Nitrogen dioxide has one fewer atom of oxygen H. Nitrogen monoxide has one fewer atom of oxygen J. Nitrogen dioxide has one more atom of nitrogen show more Which of these best describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? F. Nitrogen monoxide has one more atom of nitrogen. G. Nitrogen dioxide has one fewer atom of oxygen H. Nitrogen monoxide has one fewer atom of oxygen J. Nitrogen dioxide has one more atom of nitrogen
Chemistry
1 answer:
Pani-rosa [81]3 years ago
3 0
Nitrogen monoxide has 1 oxygen atom and
Nitrogen dioxide has 2 oxygen atoms
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As a technician in a large pharmaceutical research firm, you need to produce 250. mL of a potassium dihydrogen phosphate buffer
Reika [66]

Answer:

We will need 147.772 mL of KH2PO4 to make this solution

Explanation:

For this case we can give the following equation:

H2PO4 - ⇄ H+ + HPO42-

With following pH- equation:

pH = pKa + log [HPO42-]/[H2PO4-]

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

-0.16 =  log [HPO42-]/[H2PO4-]

10^-0.16 = [HPO42-]/[H2PO4-]

0.6918 = [HPO42-]/[H2PO4-]

Let's say the volume of HPO42-= x  then the volume of H2PO4- will be 250 mL - x

Since both have a concentration of 1M = 1 mol /L

If we plug this in the equation 0.6918 = [HPO42-]/[H2PO4-]

0.6918 = x / (250 - x)

0.6918*250 - 0.6918x = x

172.95 = 1.6918x

x = 102.228 mL

The volume of HPO42- = 102.228 mL

Then the volume of H2PO4- = 250 - 102.228 = 147.772 mL

To control this we can plug this in the pH equation

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

7.05= 7.21 + log (102.228 / 147.772) = 7.05

We will need 147.772 mL of KH2PO4 to make this solution

3 0
3 years ago
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
Please help me to do this assignment
kirill [66]

Answer:

1. Objective

2. Objective

3. Opinion

4. Objective

5. Opinion

6. Opinion

7. Opinion (I think.)

8. Opinion (I think.)

Explanation:

7 0
3 years ago
Which statement describes a chemical property of sodium
Alenkasestr [34]

Answer:

Option b is show the chemical property of sodium....

4 0
3 years ago
When liquid silver nitrate and liquid sodium chloride are combined it becomes
Maru [420]

Answer:

solid silver chloride forms along with a new liquid, sodium nitrate

Explanation:

6 0
3 years ago
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