All categories

4 years ago

What is the Ka of a 0.653 m solution of hydrocyanic acid with a ph of 5.47

Chemistry

1 answer:

vredina [299]4 years ago

5 0

Answer:

1.758 x 10⁻¹¹.

Explanation:

∵ pH = - log[H⁺].

∴ 5.47 = -log[H⁺].

log[H⁺] = -5.47.

∴ [H⁺] = 3.388 x 10⁻⁶ M.

∵ [H⁺] = √(Ka.C)

∴ [H⁺]² = Ka.C

∴ Ka = [H⁺]²/C = (3.388 x 10⁻⁶)²(0.653) = 1.758 x 10⁻¹¹.

You might be interested in

Name two elements, compounds and mixtures that are essential for living organisms. Also write why are they essential?

svlad2 [7]

elements:

calcium : for strong bones

Iron : maintaining haemoglobin for metabolism

compunds

sodium chloride : to maintain blood pressure and other life processes

Adenosine Triphosphate: for metabolism, to maintain rate of inhalation and exhalation of oxygen and to supply energy

Mixture:

I) gasoline : used as fuel

ii) cement : used in construction

5 0

3 years ago

Read 2 more answers

Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M

Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to identify the limiting reactant:

We have:

0.20 M * 50.0 mL = 10 mmol of AgNO₃

0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. That makes K₂CrO₄ the limiting reactant.

Now we calculate the mass of Ag₂CrO₄ formed, using the limiting reactant:

4 mmol K₂CrO₄ * $\frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4}$ = 1326.92 mg Ag₂CrO₄

1326.92 mg / 1000 = 1.327 g Ag₂CrO₄

7 0

3 years ago

A 25.0g sample of brass, which has a specific heat capacity of 0.375·J·g−1°C−1, is dropped into an insulated container containin

Angelina_Jolie [31]

Answer:

The equilibrium temperature of water is 25.6 °C

Explanation:

Step 1: Data given

Mass of the sample of brass = 25.0 grams

The specific heat capacity = 0.375 J/g°C

Mass of water = 250.0 grams

Temperature of water = 25.0 °C

The initial temperature of the brass is 96.7°C

Step 2: Calculate the equilibrium temperature

Heat lost = heat gained

Q(sample) = -Q(water)

Q = m*C* ΔT

m(sample)*c(sample)*ΔT(sample) = - m(water)*c(water)*ΔT(water)

⇒m(sample) = the mass of the sample of brass = 25.0 grams

⇒with c(sample) =The specific heat capacity = 0.375 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 =T2 - 96.7 °C

⇒with m(water) = the mass of the water = 250.0 grams

⇒with c(water) = the specific heat capacity = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 25.0°C

25.0 * 0.375 * (T2 - 96.7) = - 250.0 * 4.184 J/g°C * (T2 - 25.0°C)

9.375T2 - 906.56 = -1046T2 + 26150

9.375T2 + 1046T2 = 26150 + 906.56

1055.375T2 = 27056.26

T2 = 25.6 °C

The equilibrium temperature of water is 25.6 °C

4 0

3 years ago

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of

dsp73

The first thing we need to do here is to recognize the unit of molarity and the units of the given percentage of nitric acid.

Molarity is mol HNO3 / L of solution. This is our aim

The given percentage is 0.68 g HNO3/ g solution

multiplying this with density to convert g solution into mL solution and dividing with the molecular weight of HNO3 (63 g/mol) to convert g HNO3 to mol. Therefore we obtain

0.016 mol/ mL or 16.23 mol/ L (M)

6 0

4 years ago

Given the balanced equation, 2 H2 + O2 -> 2 H2O, answer the following

9966 [12]

Answer: $O_2$ is the limiting reagent

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} H_2=\frac{10.0g}{2g/mol}=5.00moles$

$\text{Moles of} O_2=\frac{20.0g}{32g/mol}=0.625moles$

The balanced chemical reaction is :

$2H_2+O_2\rightarrow 2H_2O$

According to stoichiometry :

1 moles of $O_2$ require = 2 moles of $H_2$

Thus 0.625 moles of $O_2$ will require= $\frac{2}{1}\times 0.625=1.25moles$ of $H_2$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent as it is present more than the required amount.

8 0

3 years ago