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Mekhanik [1.2K]

2 years ago

11

The acid-dissociation constant at 25.0 °c for hypochlorous acid (hclo) is 3.0 ⋅ 10−8. at equilibrium, the molarity of h3o+ in a

0.055 m solution of hclo is ________.

1 answer:

maks197457 [2]2 years ago

3 0

Hypochlorous acid is a weak acid. The $K_{a}$ value for the dissociation of HOCl is $3.0 * 10^{-8}$

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Determine how many grams of silver would be produced, if 12.83 x 10^23 atoms of copper react with an excess of silver nitrate. G

AnnyKZ [126]

1) Chemical equation

Cu + 2AgNO3 ---> Cu (NO3)2 + 2Ag

2) molar ratios

1 mol Cu: 2 moles AgNO3 : 1 mol Cu (NO3)2 : 2 mol Ag

3) Convert 12. 83 * 10^23 atoms of Cu in moles

12.83 * 10 ^ 23 atoms / (6.02 * 10^23 atoms / mol) = 2.131 mol Cu

4) Use the proportions

2.131 mol Cu * 2 mol Ag / 1 mol Cu = 4.262 mol Ag

5) Use the atomic mass of silver to convert 4.262 mol in grams

mass = number of moles * atomic mass = 4.262 mol * 107.9 g / mol = 459.9 grams

Answer: 459.9 g

5 0

2 years ago

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c

Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

$18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M$

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

$42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M$

6 0

2 years ago

A student is given a sample of CuSO4(s) that contains a solid impurity that is soluble and colorless. The student wants to deter

DochEvi [55]

Answer:

The impurity which is present in the solution of copper sulphate (CuSO4) is determined by the an instrument known as spectrophotometer.

Explanation:

Spectrophotometer is a device or an instrument which is used to determine the concentration of a chemical by measuring the detection of light intensity that is coming from the solution. If the solution of copper sulphate is checked through spectrophotometer, we can can determined or measure the amount of copper sulphate and the impurity in the solution.

3 0

2 years ago

A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),

lbvjy [14]

Answer:

[KOH] = 1.47 M

[KOH] = 1.22 m

KOH = 6.86 % m/m

Explanation:

Let's analyse the data

1.87 L is the volume of solution

Density is 1.29 g/mL → Solution density

155 g of KOH → Mass of solute

Moles of solute is (mass / molar mass) = 2.76 moles.

Molarity is mol/L → 2.76 mol / 1.87 L = 1.47 M

Let's determine, the mass of solvent.

Molality is mol of solute / 1kg of solvent

We can use density to find out the mass of solution

Mass of solution - Mass of solute = Mass of solvent

Density = Mass / volume

1.29 g/mL = Mass / 1870 mL

Notice, we had to convert L to mL, cause the units of density.

1.29 g/mL . 1870 mL = Mass → 2412.3 g

2412.3 g - 155 g = 2257.3 g of solvent

Let's convert the mass of solvent to kg

2257.3 g / 1000 = 2.25kg

2.76 mol / 2.25kg = 1.22 m (molality)

% percent by mass = mass of solute in 100g of solution.

(155 g / 2257.3 g) . 100g = 6.86 % m/m

5 0

2 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 1.50 grams of aluminum foil in a solut

Lisa [10]

Answer: d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.

Explanation: $2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of copper chloride.

According to mole concept, 1 mole of every substance weighs equal to its molar mass.

Aluminium is the limiting reagent as it limits the formation of product and copper chloride is the excess reagent as (14-7.5)=6.5 g is left as such.

Thus 54 g of of aluminium react with 270 g of copper chloride.

1.50 g of aluminium react with= $\frac{270}{54}\times 1.50=7.5 g$ of copper chloride.

3 moles of copper chloride gives 3 moles of copper.

7.5 g of copper chloride gives 7.5 g of copper.

8 0

2 years ago

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