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GaryK [48]
3 years ago
12

A solution contains 6.21 g of ethylene glycol dissolved in 25.0 g of water. If water has a boiling point elevation constant of 0

.512°C•kg/mol, what is the boiling point of the solution? (molar mass of ethylene glycol = 62.1 g/mol; boiling point of pure water = 100.00°C)
Chemistry
2 answers:
adell [148]3 years ago
7 0
The answer is D.) i just took the quiz
timurjin [86]3 years ago
5 0
Calculate first the number of moles of ethylene glycol by dividing the mass by the molar mass.
                           n = (6.21 g ethylene glycol) / 62.1 g/mol
                              n = 0.1 mol
Then, calculate the molality by dividing the number of moles by the mass of water (in kg).
                           m = 0.1 mol/ (0.025 kg) = 4m
Then, use the equation,
                       Tb,f = Tb,i + (kb)(m)
Substituting the known values,
                       Tb,f = 100°C + (0.512°C.kg/mol)(4 mol/kg)
                          <em>Tb,f = 102.048°C</em>
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Find the molarity: 9.82 grams of lead (IV) nitrate are dissolved to make 465 mL of solution.
solmaris [256]
Molarity is moles per liter. Therefore you must covert grams->moles by dividing 9.82 by the molar mass of lead IV nitrate. Then convert mL->L. Once you have those two numbers you can plug them into the formula Molarity = mol/L.
3 0
3 years ago
12.0 of .500 M NaOH neutralized 6.0 ml of HCl solution. What was the molarity of the HCl
frozen [14]

The molarity of the HCl is 1 M when 12.0 of .500 M NaOH neutralized 6.0 ml of HCl solution.

Explanation:

Data given:

molarity of the base NaOH, Mbase =0. 5 M

volume of the base NaOH, Vbase = 12 ml

volume of the acid, Vacid = 6 ml

molarity of the acid, Macid = ?

The titration formula for acid and base is given as:

Mbase Vbase = Macid Vacid

Macid =\frac{0. 5 X 12}{6}

Macid = 1 M

we can see that 1 M solution of HCl was used to neutralize the basic solution of NaOH. The volume of NaOH is 12 ml and volume of HCl used is 6ml.

8 0
3 years ago
Answer the following questions about the fermentation of glucose (C6H12O6, molar mass 180.2 g/mol)
Yuri [45]

The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal

How to determine the mole of glucose

Mass of glucose = 49 g

Molar mass of glucose = 180.2 g/mol

Mole of glucose = ?

Mole = mass / molar mass

Mole of glucose = 49 / 180.2

Mole of glucose = 0.272 mole

How to determine the energy released

C₆H₁₂O₆ →2C₂H₆O + 2CO₂  ΔH = -16 kcal/mol

From the balanced equation above,

1 mole of glucose released -16 kcal of energy

Therefore,

0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal

Thus, -4.4 Kcal were released from the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

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6 0
2 years ago
What is the [OH-] if the [H3O+] is 1 x 10 -10?
anastassius [24]

The answer is [OH⁻] = 1 x 10⁻⁴.

[OH⁻] = H₂O ÷ [H₃O⁺]

[OH⁻] = 1 x 10⁻¹⁴ / 1 x 10⁻¹⁰

[OH⁻] = 10⁻⁴

5 0
2 years ago
An object has a mass of 4 grams and a volume of 2 cm3. What is the density of the object
kolbaska11 [484]

Answer:

<h3>The answer is 2.0 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass = 4 g

volume = 2 cm³

We have

density =  \frac{4}{2}  \\

We have the final answer as

<h3>2.0 g/cm³</h3>

Hope this helps you

3 0
3 years ago
Read 2 more answers
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