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3 years ago

How many moles are there in 1.09 kg of Al(OH)3?

1 answer:

4 0

question 1
moles = mass/molar mass of Al(OH)3
convert Kg to g
that is 1.09 x 1000=1090g

moles is therefore=1090g/78(molar mass of Al(OH)3)= 13.974 moles

question 2
moles=2.55g/327.2(molar mass of Pb(CO3)2= 7.79 x 10^-3 moles
from avogadro constant
1moles=6.02 x10^23 formula units
what about 7.79 x 10 ^-3
={(7.79 x 10^-3)moles x ( 6.02 x10^23)} /1 mole=4.69 x10^21 formula units

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What will occurs during electrolysis of dilute sulphuric acid

Alexxandr [17]

Answer:

In the electrolysis of dilute sulfuric acid, which electrolysis in aqueous solution to form hydrogen ions, H⁺, and sulfur IV ions SO₄²⁻ in the presence of H⁺ and OH⁻ ions from the water molecules

At the anode

The anode, positive electrode, attracts the negative OH⁻ and SO₄²⁻ ions where the OH⁻ gives up electrons to form water molecules and oxygen as follows;

4OH⁻ → 2H₂O + O₂ + 4e⁻

At the cathode

The positive H⁺ ions from the water molecules and the acid are attracted to the cathode where they combine with 2 electrons to form hydrogen gas as follows;

2e⁻ + 2H⁺ → H₂ (gas)

Explanation:

7 0

3 years ago

H2 and O2 react to produce H2O. What is the mass of H2O produced when 0.73 g of H2 completely reacts with 3.28 g o

Slav-nsk [51]

Hydrogen reacts with oxygen based on the following equation:
2 H2 + O2 → 2 H2O

From the periodic table:
molar mass of hydrogen = 1 gram
molar mass of oxygen = 16 grams

From the balanced equation above, we can find that:
4 grams of hydrogen react with 32 grams of oxygen to produce 36 grams of water.

This means that: 0.73 grams of hydrogen require (0.73x32) / 4 = 5.84 grams of oxygen to react with.
Since only 3.28 grams of oxygen are reacting, this means that oxygen is our limiting reagent and that the reaction would stop once the amount of oxygen is consumed.

So, we will base our calculations on oxygen.
mass of water produced from 3.28 grams of oxygen can be calculated as follows:
mass of water = (3.28 x 36) / 32 = 3.69 grams

4 0

3 years ago

Which of the following reactions is balanced?

77julia77 [94]

Answer:

B. CaCl₂ + H₂CO₃ → CaCO₃ + 2HC

Explanation:

A balanced reaction has the same number of atoms in the both sides of the reaction. In the options:

A. CaCl₂ + H₂CO₃ → 2CaCO₃ + HCI

In this reaction there is 1 Ca in reactants and 2 in products -The reaction is unbalanced-

There is 1 Ca is both sides, 2Cl, 2H, 1C and 3 Oxygens -The reaction is balanced

C. CaCl₂ + 2H₂CO₃ → CaCO₃ + HCI

There is 1 Ca in both sides but 2Cl in reactants and 1 in Cl -The reaction is unbalanced-

D. 2CaCl₂ + H₂CO₃ →CaCO₃ + HCI

There are 2 Ca in reactants and 1 in Ca -The reaction is unbalanced-

3 0

3 years ago

Lists how many valence electrons are shared, gained, or lost by each atom of the 6 compounds?

Blababa [14]

The number of electrons in an atom's outermost valence shell governs its bonding behavior.

In N₂, three electrons are being shared by each nitrogen atom, making a total of 6 shared electrons.

In CCl₄, 4 electrons are being shared by each carbon atom and 1 electron is being shared by each chlorine atom

In SiO₂, 4 electrons are being shared by each silicon atom and 2 electrons are being shared by each oxygen atom.

In AlCl₃, 3 electrons are being shared by each aluminum atom and 1 electron is being shared by each Cl atom

In CaCl₂, 2 electrons are lost by the calcium atom and 1 electron is gained by each chlorine atom

In LiBr, 1 electron is lost by the lithium atom and 1 electron is gained by the bromine atom

6 0

3 years ago

Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance

ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0

3 years ago