All categories

3 years ago

Determine the mass of 6.33 mol of iron(II) nitrate.

Chemistry

1 answer:

uysha [10]3 years ago

8 0

Answer:

1822.72 g

Explanation:

Applying,

n = R.M/M.M.................. Equation 1

Where n = number of moles of iron(II) nitrate, R.M = Reacting mass of iron(II) nitrate, M.M = molar mass of Iron(II) nitrate.

Make R.M the subject of the equation

R.M = n×M.M............. Equation 2

From the question,

Given: n = 6.33 mol

Constant: M.M of iron(II) nitrate = 287.95 g/mol

Substitute these values into equation 2

R.M = 6.33(287.95)

R.M = 1822.72 g

Hence the mass of iron(II) nitrate is 1822.72 g

You might be interested in

Calculate the energy released when 24.8 g Na2O reacts in the following reaction

WARRIOR [948]

Answer:

hey you wanna get it right try this one: 48.0 kcal

was released... at constant pressure.

Explanation:

5 0

3 years ago

Helppp nowww plsss!!!

Bond [772]

Pan 4: theyre the smallest and most broken down :)

6 0

3 years ago

Using the Brønsted theory, classify the following as either an acid or a base by placing the compound in the correct bin.

guapka [62]

Answer : The Bronsted-Lowry theory was not against the Arrhenius theory, rather it was just a modification to the previous theory of acids and bases. Hydroxide ions are considered as bases because they have the tendency to accept hydrogen ions from acids and form water.

An acid was the one which produces hydrogen ions in solution because it reacts with the water molecules by giving a proton to them.

In a nutshell, he described bases as hydrogen acceptor and acids as hydrogen donors.

7 0

3 years ago

Read 2 more answers

Which of the following item(s) explain the differences between the Ka values. Choose one or more: A. The negative charge is on t

AVprozaik [17]

Answer:

D. The electron-withdrawing fluorine atoms pull electron density from the oxygen in trifluoroacetate. The negative charge is more stabilized in trifluoroacetate by this effect.

Explanation:

The structures of trifluoroacetate and acetic acid are both shown in the image attached.

The trifluoroacetate anion (CF3CO2-), just like the acetate anion has in the middle, two oxygen atoms.

However, in the trifluoroacetate anion, there are also three electronegative fluorine atoms attached to the nearby carbon atom attached to the carbonyl, and these pull some electron density through the sigma bonding network away from the oxygen atoms, thereby spreading out the negative charge further. This effect, called the "inductive effect" stabilizes the anion formed,the trifouoroacetate anion is thus more stabilized than the acetate anion.

Hence, trifluoroacetic acid is a stronger acid than acetic acid, having a pKa of -0.18.

5 0

3 years ago

Isooctane, C8H18, is the component of gasoline from which the term octane rating derives.

lina2011 [118]

Answer:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) Mass of CO₂ produced annually from this combustion of isooctane gasoline = 1.12 × 10⁵ Kg

c) CO₂ produced from the combustion of the gasoline in a year will occupy 5.632 × 10⁷ L

d) There needs to be a minimum of 1.752 × 10⁷ moles of air and 3.92 × 10⁸ L of air for the oxygen to be in excess all through the year of gasoline combustion.

Explanation:

a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O

b) C₈H₁₈ has a density of 0.792 mg/L.

Since density = mass/volume;

mass = density × volume

Mass of C₈H₁₈ with 4.6 x 10^10 L volume = 0.792 × 4.6 x 10^10 = 3.643 × 10^10 mg = 3.643 × 10⁷ g.

To obtain the mass of CO₂ produced, we need the number of moles of C₈H₁₈ that burned.

Number of moles = mass/molar mass

Molar mass of C₈H₁₈ = (8×12) + 18 = 114g/mol

Number of moles of C₈H₁₈ = (3.643 × 10⁷)/114 = (3.2 × 10⁵) moles.

From the chemical reaction,

1 mole of C₈H₁₈ burns to give 8 moles of CO₂

(3.2 × 10⁵) moles will give 8 × 3.2 × 10⁵ = (2.56 × 10⁶) moles of CO₂

Mass of CO₂ produced = number of moles × Molar mass

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ produced = 2.56 × 10⁶ × 44 = 1.12 × 10⁸ g = 1.12 × 10⁵ kg

c) 1 mole of any gas at stp occupies 22.4L

2.56 × 10⁶ moles of CO₂ will occupy 2.56 × 10⁶ × 22.4 = 5.632 × 10⁷ L

d) 1 mole of C₈H₁₈ requires 23/2 moles of O₂ for complete combustion yearly.

3.2 × 10⁵ moles would require 3.2 × 10⁵ × 23/2 = 3.68 × 10⁶ moles of O₂

O₂ makes up 21% of the air

That is,

0.21 moles of O₂ would be contained in 1 mole of air

3.68 × 10⁶ moles of O₂ would be contained in (3.68 × 10⁶ × 1)/0.21 = 1.752 × 10⁷ moles of air.

1 mole of any gas at stp occupies 22.4L

1.752 × 10⁷ of air will occupy

1.752 × 10⁷ × 22.4/1 = 3.92 × 10⁸ L of air!

3 0

3 years ago