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Firlakuza [10]
3 years ago
5

For lunch today, I ate an apple. What type of carbohydrate did I ingest?

Chemistry
1 answer:
mafiozo [28]3 years ago
8 0

Answer:

Monosaccharide

Explanation:

Apples contain high levels of fructose, which is a monosaccharide.

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Which water carries the most oxygen in streams and rivers?
lbvjy [14]

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Fassst. A or b or c or d plzzz
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C, an elliptical motion
7 0
2 years ago
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Indicate whether the following are acidic, basic, or neutral solution
goldenfox [79]
You answer this by using the pH formula and and the relation of pH and pOH, pH = -log[H+] and 14 = pH + pOH. The correct classification are as follows:

<span>A. [H2O+]=6.0x10^-12
basic

B. [H3O+]=1.4x10^-9
basic

C. [OH-]=5.0x10^-12
acidic

D. {OH-]=3.5x10^-10
acidic

Hope this answers the question.

</span>
5 0
2 years ago
Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
Eddi Din [679]

Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

8 0
2 years ago
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