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3 years ago

A buffer solution is 0.310 M in H2SO3 and 0.304 M in NaHSO3. If Ka1 forH2SO4 is 1.7e-2 , what is the pH of this buffer solution

Chemistry

1 answer:

natka813 [3]3 years ago

4 0

Answer:

pH of the buffer solution is 1.76

Explanation:

To find the pH of a buffer we can use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

<em>Where pKa is -log Ka= 1.77 And [A⁻] is molar concentration of conjugate base, NaHSO₃ and [HA] molar concentration of weak acid, H₂SO₃</em>

<em />

Replacing values of the problem:

pH = 1.77 + log [0.304M] / [0.310M]

pH = 1.76

<h3>pH of the buffer solution is 1.76</h3>

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A chemist must dilute of aqueous aluminum chloride solution until the concentration falls to . He'll do this by adding distilled

marshall27 [118]

Answer:

0.257 L

Explanation:

The values missing in the question has been assumed with common sense so that the concept could be applied

Initial volume of the AICI3 solution $=23.1 \mathrm{mL}$

Initial Molarity of the solution $=833 \mathrm{mM}$

Final molarity of the solution $=75.0 \mathrm{mM}$

Final volume of the solution $=?$

From Law of Dilution, $M_{f} V_{f}=M_{i} V_{i}$

$\Rightarrow V_{f}=\frac{M_{i} V_{i}}{M_{f}}=\frac{833 \mathrm{mM} \times 23.1 \mathrm{mL}}{75.0 \mathrm{mM}}=256.564 \mathrm{mL}=0.256564 \mathrm{L}=0.257 L$

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3 years ago

During beta-oxidation of fatty acids, ___________ is produced in peroxisomes but not in mitochondria. A) acetyl-CoA B) FADH 2 C)

stepan [7]

Answer:

The correct option is D

Explanation:

Normally, beta-oxidation of fatty acid occurs in the mitchondrial matrix, however, when the fatty acid chains are too long, the beta-oxidation occurs in the peroxisomes <u>where the oxidation is not attached to ATP synthesis but rather transferred (i.e high energy electrons are transferred) to O₂ to form hydrogen peroxide</u> (H₂O₂). This is the major difference between the beta-oxidation that occurs in the peroxisomes to that which occurs in the mitochondria.

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2 years ago

Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.

Ronch [10]

Answer : The limiting reagent is $O_2$

Solution : Given,

Moles of methane = 2.8 moles

Moles of $O_2$ = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 5 moles of $O_2$ react with $\frac{5}{2}=2.5$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is $O_2$

8 0

3 years ago

12. What is the volume of 0.07 mol of neon gas at STP?

scoundrel [369]

<h3>Answer:</h3>

2 L Ne

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Using Dimensional Analysis
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.07 mol Ne (g)

<u>Step 2: Identify Conversions</u>

STP - 22.4 L per mole

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.07 \ mol \ Ne(\frac{22.4 \ L \ Ne}{1 \ mol \ Ne})$
Multiply: $\displaystyle 1.568 \ L \ Ne$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

1.568 L Ne ≈ 2 L Ne

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2 years ago

What’s the name of this compound?

soldier1979 [14.2K]

Answer:

decane

Explanation:

decane has 10 carbon and 22hydrogen

8 0

1 year ago