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Gelneren [198K]
3 years ago
12

What is the approximate pH of a .06 M solution of CH3COOH given that Ka= 1.78*10-5

Chemistry
1 answer:
Step2247 [10]3 years ago
4 0
CH₃COOH ⇔ CH₃COO⁻ + H⁺

[CH₃COO⁻] = [H⁺] = x

K=\frac{|CH_{3}COO^{-}|*|H^{+}|}{|CH_{3}COOH|}=\frac{x^{2}}{|CH_{3}COOH|}\\\\
1,78*10^{-5}=\frac{x^{2}}{0,06} \ \ |*0,06\\\\
0,1068*10^{-5}=x^{2}\\\\
x_{1}\approx0,001 \ \land \ \ x_{2}=\approx-0,001\\\\
pH=-log|H^{+}|=-log0,001=3
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1)
vekshin1

Typically, the actual yield is lower than the theoretical yield because few reactions complete (i.e., are not 100% efficient) or because not all of the product in a reaction is collected. It is also conceivable for the real yield to exceed the theoretical yield.

What is theoretical yield ?

Theoretical yield is what would be obtained if 100% of the reaction was completed and no product was wasted in any way; however, procedures are not this efficient. Everything is dependent on the experimenter's precision and the reaction. Some reactions are reversible, meaning that less than 100% of the reaction proceeds to completion; some reactions do not go to completion because they require a large amount of energy or additional time, for example. Substances in the reaction may be lost throughout the method or difficult to separate from other sections of the experiment. In chemistry, some degree of inaccuracy is always expected.

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5 0
1 year ago
How many chlorine molecules are in 35.5g of chlorine gas
Karo-lina-s [1.5K]
Since chlorine is one of the 7 diatomic elements we know that chlorine appears as Cl₂ gas naturally.  That means that the molar mass of a chlorine gas is 70.9g/mol.  That being said, first you need to find the number of moles of chlorine gas that are present in a 35.5g sample.  To do this divide 35.5g by the molar mass of chlorine gas (70.9g/mol) to get 0.501mol of chlorine.  Then you have to multiply 0.501mol by 6.02×10²³ to get the number of chlorine gas molecules.  Therefore 3.01×10²³ molecules of chlorine gas are present in a 35.5g sample.

I hope that helps.  Let me know in the comments if anything is unclear.
7 0
3 years ago
Someone please help me , number 14
MrRa [10]
We know, 

PV=nRT
203*20= n*0.0821*373
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6 0
3 years ago
Which of the following species cannot act as a Lewis base?
gayaneshka [121]

Answer:

NH4+

Explanation:

A Lewis base is a substance with the ability to share or give pairs of electrons.

The NH4 + ammonium ion donates a proton, therefore the ammonium ion is a Lewis acid.

6 0
3 years ago
A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of
Leviafan [203]

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

7 0
2 years ago
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