Please help me due tomorrow

Calculate the Ka for a 0.3 M solution of HA (unknown weak acid) if the pH = 3.65. The reaction can be modelled as HA (aq) + H2O

Levart [38]

The Ka : 1.671 x 10⁻⁷

<h3>Further explanation</h3>

Given

Reaction

HA (aq) + H2O (l) ←→ A- (aq) + H3O+ (aq).

0.3 M HA

pH = 3.65

Required

Ka

Solution

pH = - log [H3O+]

$\tt [H_3O^+]=10^{-3.65}=2.239\times 10^{-4}$

ICE method :

HA (aq) ←→ A- (aq) + H3O+ (aq).

0.3 0 0

2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴

0.3-2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴

$\tt Ka=\dfrac{[H_3O^+][A^-]}{[HA]}\\\\Ka=\dfrac{(2.239.10^{-4}){^2}}{0.3-2.239.10^{-4}}\\\\Ka=1.671\times 10^{-7}$

5 0

3 years ago

If i initially have a gas with a pressure of 84 kpa and a temperature of 350 c and i heat it an additional 230 degrees, what wil

Ganezh [65]

Answer:

67.824

Explanation: You want to use the combined gas law equation (P1*V1)/(n1*T1)=(P2*V2)/(n2*T2). So first cross out what remains constant, so volume(V) and I assume moles (since it was not mentioned as a change). Then you can solve algebraically for the answer!

Hope this helped!

5 0

3 years ago

From the list of substances, identify each as a strong acid, strong base, or neither a strong acid nor a strong base.

Dafna11 [192]

Answer:

HNO3 is a strong acid

HCl is also a strong acid

NaCl is a salt so it is neither a strong acid nor a strong base

Ca(OH)3 is a strong base

Explanation:

8 0

3 years ago

Need help with question 1 & 2

zhannawk [14.2K]

For question 1:

A. Carbon, hydrogen, oxygen

Calcium, carbon, oxygen

Carbon, hydrogen

Carbon, hydrogen, oxygen

Silicon, oxygen

B. Carbon: 12, hydrogen: 22, oxygen: 11

Calcium: 1, Carbon: 1, oxygen: 3

Carbon:1, Hydrogen: 4

Carbon: 3, hydrogen: 8, Oxygen: 1

Silicon: 1 oxygen: 2

C. Table sugar: 45

Marble: 5

Natural gas: 5

Rubbing alcohol: 12

Glass: 3

For question 2: N2O

6 0

3 years ago

A pharmacy intern is asked to prepare 3 L of a 30% w/v solution. T he pharmacy stocks the active ingredient in 8-ounce bottles o

MariettaO [177]

<u>Answer:</u> The number of bottles that will be needed are 6

<u>Explanation:</u>

We are given:

Amount of solution, the intern is asked to prepare = 3 L = 3000 mL (Conversion factor: 1 L = 1000 mL)

Strength of solution needed = 30 % (w/v)

This means that in 100 mL of solution, the solute present is 30 grams

So, in 3000 mL of solution, the solute present will be = $\frac{30}{100}\times 3000=900g$

Active ingredient present in 1 bottle = 8 ounce of 70 % (w/v)

<u>Conversion factor used:</u> 1 ounce = 29.57 mL

So, $8ounce\times \frac{29.57mL}{1ounce}=236.6mL$

Amount of active ingredient present in 1 bottle = $236.6\times \frac{70}{100}=165.6g$

To calculate the number of bottles, we need to divide the total amount of solution needed by the amount of active ingredient present in 1 bottle, we get:

$\text{Number of bottles}=\frac{\text{Amount of solution to prepare}}{\text{Amount of active ingredient in 1 bottle}}$

Putting values in above equation, we get:

$\text{Number of bottles}=\frac{900g}{165.6g}\\\\\text{Number of bottles}=5.43\approx 6$

Hence, the number of bottles that will be needed are 6

6 0

3 years ago