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pochemuha
3 years ago
11

Please help me with chemistry

Chemistry
1 answer:
Murljashka [212]3 years ago
7 0

Answer:

60.052 g/mol (molar mass of vinegar)

84.007g/mol (baking soda )

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Phosphate buffers are commonly used in biological research. If a small amount of strong acid is added to a buffer solution that
lozanna [386]

Answer:

A) [H3PO4] will increase, [KH2PO4] will decrease, and pH will slightly decrease.

Explanation:

A buffer is a solution which resists changes to its pH when a small amount of acid or base is added to it.

Buffers consist of a weak acid (HA) and its conjugate base (A–) or a weak base and its conjugate acid. Weak acids and bases do not completely dissociate in water, and instead exist in solution as an equilibrium of dissociated and undissociated species. When a small quantity of a strong acid is added to a buffer solution, the conjugate base, A-, reacts with the hydrogen ions from the added acid to form the weak acid and a salt thereby removing the extra hydrogen ions from the solution and keeping the pH of the solution fairly constant. On the other hand, if a small quantity of a strong base is added to the buffer solution, the weak acid dissociates further to release hydrogen ions which then react with the hydroxide ions of the added base to form water and the conjugate base.

For example, if a small amount of strong acid is added to a buffer solution that is 0.700 M H3PO4 and 0.700 M KH2PO4, the following reaction is obtained:

KH₂PO₄ + H+ ----> K+ + H₃PO₄

Therefore, [H₃PO₄] will increase, [KH₂PO₄] will decrease, and pH will slightly decrease.:

4 0
3 years ago
What energy conversion in your body enables you to get out of bed in the morning
Vadim26 [7]
<span> The thermal </span>energy<span> is being </span>converted<span> to light </span>energy,<span> you are using electrical </span>energy<span>. </span>Your body<span> relies on the chemical </span><span>energy</span>
5 0
3 years ago
Calculate the pH after 10 mL of 1.0 M sodium hydroxide is added to 60 mL of 0.5M acetic acid.
baherus [9]

Explanation:

It is given that the total volume is (10 mL + 60 mL) = 70 mL.

Also, it is known that M_{1}V_{1} = M_{2}V_{2}

Where,    V_{1} = total volume

               V_{2} = initial volume

Therefore, new concentration of CH_{3}COOH = \frac{M_{2}V_{2}}{V_{1}}

                                        = \frac{60 \times 0.5}{70}

                                        = 0.43 M

New concentration of NaOH = \frac{M_{2}V_{2}}{V_{1}}

                                               = \frac{10 \times 1.0}{70}

                                               = 0.14 M

So, the given reaction will be as follows.

              CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O

Initial:             0.43          0.14                     0

Change:          -0.14        -0.14                    0.14

Equilibrium:    0.29          0                       0.14

As it is known that value of pK_{a} = 4.74

Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.

           pH = pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

                 = 4.74 + log \frac{0.14}{0.29}

                 = 4.74 + (-0.316)

                 = 4.42

Therefore, we can conclude that the pH of given reaction is 4.42.

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