All categories

3 years ago

20. The mass of one mole of lead (Pb) atoms is 207.2 g. Use a proportion to calculate the n

Chemistry

1 answer:

AnnZ [28]3 years ago

8 0

Answer:

a) atoms Pb = 4.3595 E22 amu

b) g = ( 6.022 E23 amu ) × ( Mw )

Explanation:

m Pb = 207.2 g / mol Pb

sample:

∴ m = 15.00 g

a) atoms Pb = ?

1 mol ≡ 6.022 E23 amu

⇒ atoms Pb = (15.00g Pb)(mol Pb/207.2 g)(6.022 E23 amu/mol)

⇒ atoms Pb = 4.3595 E22 amu

b) relationship:

∴ 1 mol (n) ≡ 6.022 E23 amu......(1)

∴ mass (g) ≡ (Mw)×(n)

⇒ n = g / Mw..........(2)

∴ Mw: molecular weight

(1) = (2):

⇒ g / Mw ≡ 6.022 E23 amu

⇒ g ≡ ( 6.022 E23 amu ) × ( Mw )

You might be interested in

The three particles that make up atoms are

Anni [7]

A. Protons neutrons and electrons.

Haha those three make up a simple Atom.

6 0

3 years ago

Read 2 more answers

What percentage of radioactive substance remains after two half-lives

Ksju [112]

Answer:

After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.

Explanation:

3 0

2 years ago

Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of hcl with 1,3-pentadiene

GrogVix [38]

1,3-pentadiene has two double bonds which are conjugated, which undergo electrophilic addition reaction on reacting with $HCl$ .

The structure of 1,3-pentadiene is shown in the image.

When strong acid such as $HCl$ reacts with 1,3-pentadiene, the electrophilic addition reaction can occur either on double bond at 1,2-position or at 3,4-position. The reaction that occurs is shown in the image.

3 0

3 years ago

How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward

fomenos

Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

NH3 + H2O(l) ⇄ NH4+ + OH-

Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

<em>And as </em>[NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

pH:

As pOH = -log [OH-]

pOH = 2.48

pH = 14 - pOH

5 0

2 years ago

Urea, (NH2)2CO, is a product of metabolism of proteins. An aqueous solution is 37.2% urea by mass and has a density of 1.032 g/m

Feliz [49]

Answer:

The molarity of urea in this solution is 6.39 M.

Explanation:

Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is

$molarity = moles of solute ÷ liters of solution$

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

Our first step is to calculate the moles of urea in 100 grams of the solution,

using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is

60.06 g/mol ÷ 37.2 g = 0.619 mol

Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.

1.032 g/mL ÷ 100 g = 96.9 mL

This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.

0.619 mol/96.9 mL × 1000 mL= 6.39 M

Therefore, the molarity of the solution is 6.39 M.

4 0

3 years ago