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3 years ago

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Which one of these salts will form a basic solution upon dissolving in water?

1 answer:

daser333 [38]3 years ago

6 0

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Which acid is almost completely ionized in a dilute solution at 298k?

Alex777 [14]

Considering choices;
a. CH3COOH
b. H2S
c. H2PO4
d. HNO3
The answer would be HNO3; this is because it the strongest acid. Strong acids are acids that are fully ionized while weak acids are only partially ionized. At the same concentrations, strong acids have a higher concentration of hydrogen than weak acids. Examples of strong acids include; hydrocloric acid and nitric acid, while weak acids include; ethanoic acids.

5 0

4 years ago

A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was dil

Masteriza [31]

Answer : The value of $pK_a$ of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

$\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}$

$\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol$

Now we have to calculate the value of $pK_a$ of the weak acid.

The equilibrium chemical reaction is:

$HA+KOH\rightleftharpoons HK+H_2O$

Initial moles 0.25 0.03 0

At eqm. (0.25-0.03) 0.03 0.03

= 0.22

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[HK]}{[HA]}$

Now put all the given values in this expression, we get:

$3.85=pK_a+\log (\frac{0.03}{0.22})$

$pK_a=4.72$

Therefore, the value of $pK_a$ of the weak acid is, 4.72

7 0

3 years ago

Who wants to hang out?

vovangra [49]

What’s the question?

7 0

3 years ago

A balloon filled with air has a volume of 0.500 L at 25.0 ºC and an internal pressure of 1.20 atm. What will the volume of the b

Alex_Xolod [135]

Answer:

0.441 L

Explanation:

At constant pressure, Using Charle's law :-

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,

V₁ = 0.500 L

V₂ = ?

T₁ = 25.0 °C

T₂ = -10.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25 + 273.15) K = 298.15 K

T₂ = (-10 + 273.15) K = 263.15 K

Using above equation as:

$\frac{0.500}{298.15}=\frac{V_2}{263.15}$

$V_2=\frac{0.500\cdot \:263.15}{298.15}$

<u>New volume = 0.441 L</u>

4 0

3 years ago

A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou

Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

3 0

4 years ago