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Reil [10]
3 years ago
15

Explaination on vitamin B1​

Chemistry
1 answer:
TiliK225 [7]3 years ago
6 0

Answer:

Vitamin B1 helps some enzymes work properly, helps break down sugars in the diet, and keeps nerves and the heart healthy. It is found in pork, organ meats, peas, beans, nuts, and whole grains. Vitamin

<em>hop</em><em>e</em><em> this</em><em> answer</em><em> correct</em><em> </em><em>(</em><em>^</em><em>^</em><em>)</em><em>.</em><em>.</em><em>.</em><em>.</em>

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One mole of aluminum atoms has a mass of 27 grams. How many grams are in 9.0 moles of aluminum?
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27*9=243 if one mole is equal to 27 grams times that by 9
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Clouds are named for 3 different attributes including shape, _____________ &amp; precipitation. *
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3 years ago
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How many pi bonds are in the oxalate ion (C2O42-)?
suter [353]

Answer:there are 13

Explanation:

4 0
3 years ago
Valeric acid, HC5H9O2 (Ka = 1.5 ✕ 10−5), is used in the manufacture of magnesium valerate, a nerve-calming agent. What is the hy
Studentka2010 [4]

Answer:

[H^+]=0.00332M

Explanation:

Hello,

In this case, considering the dissociation of valeric acid as:

HC_5H_9O_2 \rightleftharpoons C_5H_9O_2 ^-+H^+

Its corresponding law of mass action is:

Ka=\frac{[H^+][C_5H_9O_2^-]}{[HC_5H_9O_2]}

Now, by means of the change x due to dissociation, it becomes:

Ka=\frac{(x)(x)}{0.737-x}=1.5x10^{-5}

Solving for x we obtain:

x=0.00332M

Thus, since the concentration of hydronium equals x, the answer is:

[H^+]=x=0.00332M

Best regards.

7 0
3 years ago
What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250
Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons \text{H}^{+} than hydroxide ions \text{OH}^{-}. The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that x \; \text{L} of the 0.307 \text{mol} \cdot \text{dm}^{-3} sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, x is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain 0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x moles of acetic acid and 0.307 \; x moles of acetate ions.

Let \text{HAc} denotes an acetic acid molecule and \text{Ac}^{-} denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of 10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}. There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}

By definition:

\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)

The question states that

\text{K}_{a} = 1.75 \times 10^{-5}

such that

10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241

Thus it takes 0.0241 \; \text{L} of sodium hydroxide to produce this buffer solution.

6 0
3 years ago
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