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Dahasolnce [82]
3 years ago
14

n today's experiment, Solutions A and B are prepared as follows. Solution A: Solution B: 2.0 mL of 3.00 x 10-4 M bromcresol gree

n 2.0 mL of 3.00 x 10-4 M bromcresol green 5.0 mL of 1.60 M acetic acid (HAc) 2.0 mL of 0.160 M sodium acetate (NaAc) 2.0 mL of 0.200 M KCl diluted to a total volume of 50 mL diluted to a total volume of 50 mL How many mL of Solution A must be added to Solution B to give a buffer that is equimolar in HAc and Ac-
Chemistry
1 answer:
SOVA2 [1]3 years ago
4 0

Answer:

2 mL of Solution A must be added to Solution B to give a buffer that is equimolar.

Explanation:

 Given the data in the question;

First we determine the number of sodium acetate;  

⇒ molarity × volume ( L )

⇒ 0.16 × 2.0 mL

⇒ 0.16 × 0.002 L

⇒ 0.00032

Now, Molarity of sodium acetate = moles / Volume(L)

⇒ ( 0.00032 / 50 ) × 1000

⇒ 0.0064

Since number of moles of acetic acid that should be added tp make equimolar solution is 0.00032

and Molarity of acetic acid is 0.16 molL⁻¹

Let X represent the volume that should be added.

so;

Molarity = Moles / Volume (L)

we substitute

0.16 = (0.00032  / X) × 1000

0.16 = 32 / X

X = 0.32 / 0.16

X = 2 mL

Therefore, 2 mL of Solution A must be added to Solution B to give a buffer that is equimolar.

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The best solution to deforestation is to curb the felling of trees by enforcing a series of rules and laws to govern it. Deforestation in the current scenario may have reduced; however, it would be too early to assume. The money-churner nature of forest resources can be tempting enough for deforestation to continue.

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2 years ago
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a brick measures of 13 inches i length 10 iches in height and 4 inches in width what is the vlume of the brick
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4 0
2 years ago
Indicate which of the following criteria are important for the selection of a buffer to use in an in vitro biochemical reaction.
JulsSmile [24]

Answer:

The correct answer is: d. The pKa of the chosen buffer should be close to the optimal pH for the biochemical reaction.

Explanation:

The buffer resist or maintain the change in pH in case of Acid or basic addition to the solution. The buffer capacity should be within one or two pH units when compared to the optimal pH.

Thus it is important to select a buffer with pKa close to the optimum pH of the reaction because the ability for the buffer to maintain the pH is is great at the pH close to pKa.

7 0
2 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
2 years ago
(2pts) During the Purification of Lactate Dehydrogenase (LDH) experiment, you will need 50ml of buffer A150. Buffer A150 is 30mM
torisob [31]

Answer:

The answer is "20 \ mL"

Explanation:

Given:

Molarity= number of moles

because it is 1 Liter

\to \frac{0.03\ moles}{1.5 moles}=0.02\ L= 20 \ mL \ of\  Tris\\\\

therefore,

it takes 20 mL of Tris.

\to \frac{0.150 \ moles}{5\ moles} =0.03\ L\\\\

                     = 30 \ mL \ of\ Nacl

So, take 20 \ mL\ of\ NaCl.

6 0
3 years ago
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