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3 years ago

Calculate the oxidation numbers of the elements in this equation 2 Al2O3 → 4 Al + 3 O2

Chemistry

1 answer:

riadik2000 [5.3K]3 years ago

5 0

Explanation:

Answer

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The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.

Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O ) = 0

2× ( Oxidation number of Al) +3(−2)=0

2× ( oxidation number of Al) +6

∴ Oxidation number of Al =+3

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Which process of the water cycle occur by releasing energy?

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Condensation occurs through the release of energy

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To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

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Answer: The number of moles of weak acid is $4.24\times 10^{-3}$ moles.

Explanation:

To calculate the moles of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

$0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol$

The chemical reaction of weak monoprotic acid and KOH follows the equation:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, $4.24\times 10^{-3}mol$ of KOH will react with = $\frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol$ of weak monoprotic acid.

Hence, the number of moles of weak acid is $4.24\times 10^{-3}$ moles.

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3 years ago

Martin slowly pours an unknown liquid into a container that originally had some water in it. He is measuring the temperature in

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B. a chemical change

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2 years ago

Read 2 more answers

If a gas is initially at a pressure of 9 atm, a volume of 21 liters, and a temperature of 253 K, and then the pressure is raised

alex41 [277]

Answer:

15.04 mL

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 21 L

V₂ = ?

P₁ = 9 atm

P₂ = 15 atm

T₁ = 253 K

T₂ = 302 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{9}\times {21}}{253}=\frac {{15}\times {V_2}}{302}$

Solving for V₂ , we get:

V₂ = 15.04 mL

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2 years ago

A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati

Katen [24]

Answer:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

Explanation:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

$Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}$

The pKa is

$-Log (Ka) = -Log (6.2x10^{-8}) = 7.2$

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.

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3 years ago