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3 years ago

Determine the molality of an aqueous solution that is 15.5 percent urea by mass.

1 answer:

3 0

15.5% by mass is equivalent 15.5 g urea in 100 g solution or 155 g urea in 1 kg solution. 

we know that molality = moles solute / kg solvent

moles solute = 155 g / 60 g/mol = 2.58 moles urea

Since there are 155 g urea in 1000g solution, hence the solvent is 845 g or 0.845 kg

So:
molality = 2.58 / 0.845 = 3.06 m

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If 35 ml of 6.0 m h2so4 was spilled, calculate the minimum mass of nahco3 that must be added to the spill to neutralize the acid

belka [17]

The balanced equation between the $H_2SO_4$ and $NaHCO_3$ is:

$H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O$

Formula of molarity is:

$Molarity = \frac{Moles of solute}{Volume of solution in Liters}$

$Molarity = 6.0 M, Volume = 35 mL = 0.035 L$

Substituting the values,

$6 = \frac{Moles of solute}{0.035}$

$Moles of solute = 0.035 L\times 6 mol/L = 0.21 mole$

So, number of moles of $H_2SO_4$ is $0.21 mole$ .

From the balanced equation it is clear that for 1 mole of $H_2SO_4$ , 2 moles of $NaHCO_3$ are required.

Hence, 0.21 mole of $H_2SO_4$ = $2\times 0.21 mole = 0.42 mole of NaHCO_3$

Molar mass of $NaHCO_3$ = $84.007 g/mol$

So, the mass of $NaHCO_3$ = $84.007 g/mol \times 0.42 mol = 35.283 g$

6 0

3 years ago

Which set of numbers gives the correct possible values of 1 for n = 3?

Allushta [10]

Answer:

A. 0,1,2

Explanation:

took a test on it and this was the correct answer

7 0

3 years ago

Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make ___ mol ammonium sulfate

Bingel [31]

Answer:

1) Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.

2) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.

Explanation:

Firstly, we should balance the equation of heptane combustion.
We can balance the equation by applying the conservation of mass to the equation.

The balanced equation is: 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.

This means that every 2.0 moles of ammonium hydroxide (NH₄OH) will produce 1.0 mole of ammonium sulfate (NH₄)₂SO₄ when it is neutralized by sulfuric acid.

We need to calculate the no. of moles in 11.0 g of ammonium hydroxide that is neutralized using the relation: n = mass/molar mass.

n of 11.0 g of ammonium hydroxide (NH₄OH) = mass/molar mass = (11.0 g)/(35.04 g/mol) = 0.314 mol.

Using cross multiplication:

2.0 moles of NH₄OH make → 1.0 mole of (NH₄)₂SO₄.

0.314 mol of NH₄OH make → ??? moles of (NH₄)₂SO₄.

∴ The no. of moles of (NH₄)₂SO₄ that will be made from neutralizing (11.0 g) of NH₄OH = (0.314 mol)(1.0 mol)/(2.0 mol) = 0.157 mol.

∴ Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.

3 0

3 years ago

Hurry plz

erik [133]

Answer:

1) The correct step in the scientific method that Victor did is Construct a hypothesis.

2) Given mass and density, volume is calculated as mass divided by density.

Explanation:

1) Before doing the assay and make a graph with the results obtained, Victor should think what he wants to prove, so he should make a hypoythesis to test with the assay.

2) The formula of density is

density = mass/volume ⇒ density x volume = mass ⇒ volume = mass/density.

4 0

3 years ago

Read 2 more answers

What is the pH of a 5.09 x 10-5 M solution of NaOH?

Varvara68 [4.7K]

Answer: The pH of the solution is 9.71

Explanation:

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions.

We are given:

pOH of the solution = 7.2