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vesna_86 [32]
3 years ago
13

Drinking water may contain a low concentration of lead ion (Pb2 ) due to corrosion of old lead pipes. The EPA has determined tha

t the maximum safe level of lead ion in water is 15 ppb. Suppose a sample of tap water was determined to have a lead ion concentration of 0.0080 ppm. Assume the density of the solution is 1.00 g/mL. How many moles of lead ion are in 100.0 mL of the water
Chemistry
1 answer:
Lynna [10]3 years ago
4 0

Answer:

3.861x10⁻⁹ mol Pb⁺²

Explanation:

We can <u>define ppm as mg of Pb²⁺ per liter of water</u>.

We<u> calculate the mass of lead ion in 100 mL of water</u>:

  • 100.0 mL ⇒ 100.0 / 1000 = 0.100 L
  • 0.100 L * 0.0080 ppm = 8x10⁻⁴ mg Pb⁺²

Now we <u>convert mass of lead to moles</u>, using its molar mass:

  • 8x10⁻⁴ mg ⇒ 8x10⁻⁴ / 1000 = 8x10⁻⁷ g
  • 8x10⁻⁷ g Pb²⁺ ÷ 207.2 g/mol = 3.861x10⁻⁹ mol Pb⁺²
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PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PC
Dmitry_Shevchenko [17]

Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

                PCl₃(g)        +        Cl₂(g)     ⇋        PCl₅(g)

Initially     0.24 M                 1.50M                 0.12 M

React           x                           x                         x

Some amount of compound has reacted during the process.

In equilibrium we have

              0.24 - x                  1.50 - x                  0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)          

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

5 0
3 years ago
Calculate the activation energy in kJ/mol for a reaction that takes place at 298K with a rate constant of 6.87 x 10-2 and the sa
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3 0
2 years ago
After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of
SIZIF [17.4K]

Answer:

b) The dehydrated sample absorbed moisture after heating

Explanation:

a) Strong initial heating caused some of the hydrate sample to splatter out.

This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).

b) The dehydrated sample absorbed moisture after heating.

Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.

c) The amount of the hydrate sample used was too small.

It will create some errors but they do not create a difference of 13% difference as stated in the problem.

d) The crucible was not heated to constant mass before use.

Here the error is small.

e) Excess heating caused the dehydrated sample to decompose.

Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.

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3 years ago
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Answer:

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4 0
2 years ago
In the reaction below, if 5.71 g of sulfur is reacted with 10.0 g of oxygen, how many grams of sulfur trioxide will be produced?
yawa3891 [41]

Answer:

14.3 g SO₃

Explanation:

2S + 3O₂ → 2SO₃

First, find the limiting reactant.  To do that, calculate the mass of oxygen needed to react with all the sulfur.

5.71 g S × (1 mol S / 32 g S) = 0.178 mol S

0.178 mol S × (3 mol O₂ / 2 mol S) = 0.268 mol O₂

0.268 mol O₂ × (32 g O₂ / mol O₂) = 8.57 g O₂

There are 10.0 g of O₂, so there's enough oxygen.  The limiting reactant is therefore sulfur.

Use the mass of sulfur to calculate the mass of sulfur trioxide.

5.71 g S × (1 mol S / 32 g S) = 0.178 mol S

0.178 mol S × (2 mol SO₃ / 2 mol S) = 0.178 mol SO₃

0.178 mol SO₃ × (80 g SO₃ / mol SO₃) = 14.3 g SO₃

3 0
3 years ago
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