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sweet-ann [11.9K]
3 years ago
9

Calculate the number of Carbon atoms in 1.50g of C6H12O6

Chemistry
2 answers:
Nostrana [21]3 years ago
5 0

Answer:

Number of C-atoms =  3.01 *10^22

Explanation:

<u>Step 1: </u>Data given

Mass of glucose = 1.50 grams

Molar mass of Glucose = 180.156 g/mol

Molar mass of carbon = 12.01 g/mol

<u>Step 2: </u>Calculate number of moles glucose

Number of moles glucose = mass glucose / molar mass glucose

Moles glucose = 1.50 grams / 180.156 g/mol

Moles glucose = 0.00833 moles

<u>Step 3:</u> Calculate formula units of glucose

0.00833 moles glucose * 6.02 *10^23 = 5.01 *10^21

<u>Step 4:</u> Calculate number of carbon atoms

Glucose has 6 C'atoms in it's molecular formule

Number of C-atoms = 5.01 *10^21 * 6 = 3.01 *10^22

Whitepunk [10]3 years ago
4 0

Answer:

3.01 ·10↑22

Explanation:

First you want to convert the grams of Glucose to moles of Glucose.

\frac{1.5 grams of glucose}{180.15588grams/molesglucose} =.008326 Moles of Glucose

Next find the formula units of glucose.

.008326Moles of Glucose · 6.022 · 10↑23Forumula Units*Moles↑-1 =

5.01 ·10↑21 Formula Units of Glucose

Now multiply the formula units of glucose by the amount of each element in the molecule.

So for Carbon:

6carbon · 5.01 · 10↑21 = 3.01 · 10↑22

(please mark as Brainliest)

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An analytical chemist is titrating of a solution of ethylamine with a solution of . The of ethylamine is . Calculate the pH of t
Elenna [48]

Answer:

pH=11.

Explanation:

Hello!

In this case, since the data is not given, it is possible to use a similar problem like:

"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"

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C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol

Thus, the concentration of ethylamine in solution is:

[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M

Now, we can also infer that some salt is formed, and has the following concentration:

[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M

Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:

pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0

Finally, the pH turns out to be:

pH=14-pOH=14-3\\\\pH=11

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.

Best regards!

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