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8_murik_8 [283]
3 years ago
13

The density of a 22.0% by mass ethylene glycol (c2h6o2) solution in water is 1.04 g/ml . part a find the molarity of the solutio

n.
Chemistry
1 answer:
julsineya [31]3 years ago
4 0

Answer:- 3.68M.

Solution:- We have a 22.0% by mass solution of ethylene glycol and it's density is 1.04 gram per mL. It ask to calculate the molarity of the solution. We know that molarity is moles of solute per liter of solution. So, we need to figure out the moles of ethylene glycol and volume of solution.

Let's say we have 100 grams of the solution. Then mass of ethylene glycol would be 22.0 grams. Molar mass of ethylene glycol is 62.07 gram per mol.

Let's calculate it's moles first:

22.0gC_2H_6O_2(\frac{1mol}{62.07g})

= 0.354molC_2H_6O_2

From mass and density we calculate the volume of the solution and convert it to liters as:

100g(\frac{1mL}{1.04g})(\frac{1L}{1000mL})

= 0.0962 L

molarity=\frac{0.354mol}{0.0962L}

= 3.68M

So, the molarity of ethylene glycol solution is 3.68M.

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Answer is: dispersion forces.

The London dispersion force is the weakest intermolecular force.

Dispersion force is also called an induced dipole-induced dipole attraction.

The London dispersion force (intermolecular force) is a temporary attractive force between molecules.

The dipole beetween iodine and bromine is weak.

8 0
3 years ago
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Calculate the mass percent for all components in a solution containing the following. 0.350 Kg of water, 5.4 moles of ammonia an
Alina [70]

Answer:

  • % Water = 54.96%
  • % Ammonia = 0.14%
  • % Cobalt (II) Nitrate = 30.62%

Explanation:

To calculate mass percent, first we need to <u>calculate the total mass of the mixture</u>:

  • Mass Water ⇒ 0.350 kg Water = 350 g water
  • Mass Ammonia⇒We use ammonia's molar mass⇒5.4 mol * 17 g/mol =  91.8 g
  • Mass cobalt (II) nitrate ⇒ 195.0 g

Total Mass = Mass Water + Mass Ammonia + Mass Cobalt Nitrate

  • Total Mass = 350 g+ 91.8 g+ 195 g = 636.8 g

To calculate each component's mass percent, we divide its mass by the total mass and multiply by 100:

  • % Water ⇒ 350/636.8 * 100% = 54.96%
  • % Ammonia ⇒ 91.8/636.8 * 100% = 0.14%
  • % Cobalt (II) Nitrate ⇒ 195/636.8 * 100% = 30.62%
8 0
4 years ago
A chemist dissolves of pure potassium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The t
Leya [2.2K]

Answer:

12.99

Explanation:

<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>

Step 1: Given data

  • Mass of KOH: 716. mg (0.716 g)
  • Volume of the solution: 130. mL (0.130 L)

Step 2: Calculate the moles corresponding to 0.716 g of KOH

The molar mass of KOH is 56.11 g/mol.

0.716 g × 1 mol/56.11 g = 0.0128 mol

Step 3: Calculate the molar concentration of KOH

[KOH] = 0.0128 mol/0.130 L = 0.0985 M

Step 4: Write the ionization reaction of KOH

KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)

The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M

Step 5: Calculate the pOH

We will use the following expression.

pOH = -log [OH⁻] = -log 0.0985 = 1.01

Step 6: Calculate the pH

We will use the following expression.

pH + pOH = 14

pH = 14 - pOH = 14 -1.01 = 12.99

8 0
3 years ago
Which statement describes a strong acid?
masya89 [10]

Answer:

A

Explanation:

B describes a strong base, C just isn't true there are only 7 strong acids, D describes a weak acid

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You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with K
seropon [69]

Answer:

HA +  KOH  →  KA  +  H₂O

Explanation:

The unknown solid acid in water can release its proton as this:

HA  +  H₂O  →  H₃O⁺  +  A⁻

As we have the anion A⁻, when it bonded to the cation K⁺, salt can be generated, so the reaction of HA and KOH must be a neutralization one, where you form water and a salt

HA +  KOH  →  KA  +  H₂O

It is a neutralization reaction because H⁺ from the acid and OH⁻ from the base can be neutralized as water

7 0
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