What happens to the density of the air in the balloon as it is heated?

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

2 years ago

Use this information to determine the number of valence electrons in the atoms. Which of the following correctly compares the st

vlabodo [156]

The question is incomplete. The complete question is :

The table compares the number of electrons in two unknown neutral atoms.

Comparison of Electrons

Atom Number of Electrons

A 9

D 11

Use this information to determine the number of valence electrons in the atoms. Which of the following correctly compares the stability of the two atoms?

A. Both are unreactive.

B. Both are highly reactive.

C. A is unreactive and D is reactive.

D. A is reactive and D is unreactive.

Answer:

B. Both are highly reactive.

Explanation:

Atomic number of electrons given is 9 and 11, it means that atom A is fluorine (F) and atom D is Sodium (Na).

In order to gain stability 8 electrons are required in outer shell. So atoms either gain or lose electrons in its outer shell to become more reactive.

Sodium has electronic configuration (2,8,1) and easily loses one electron and become (2, 8) which is highly reactive.

Fluorine has electronic configuration (2, 7) and can gain easily an electron and becomes reactive to form (2, 8).

So, both they both attain stability by losing and gaining only one electron.

Hence, they both are highly reactive and the correct option is B.

8 0

3 years ago

What feature of single nucleotides provides the energy needed for polymerization when nucleic acids are formed?

lukranit [14]

When the phosphate groups of the single nucleotides combine to form the backbone of the nucleic acid, energy is released. This energy is used for polymerization.

4 0

3 years ago

Which of the following best describes the first step in the formation of a hurricane?

irinina [24]

I think the answer is B...

I am not sure

5 0

3 years ago

Read 2 more answers

Can you help me balance this

Zepler [3.9K]

Balanced equation

Mg₃(PO₄)₂ + 2Al(NO₃)₃ → 3Mg(NO₃)₂ + 2AlPO₄

<h3>Further explanation</h3>

Given

Reaction

Mg3(PO4)2 + Al(NO3)3 → Mg(NO3)2 + AlPO4

Required

Balanced equation

Solution

Give a coefficient :

Mg₃(PO₄)₂ + aAl(NO₃)₃ → bMg(NO₃)₂ + cAlPO₄

Make an equation :

Mg, left = 3, right=b⇒b=3

P, left=2, right=c⇒c=2

Al, left=a, right=c⇒a=c⇒a=2

The reaction becomes :

Mg₃(PO₄)₂ + 2Al(NO₃)₃ → 3Mg(NO₃)₂ + 2AlPO₄

7 0

3 years ago