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3 years ago

Which of the following equations does not represent a redox reaction?

Chemistry

1 answer:

bija089 [108]3 years ago

4 0

Answer:

a. NH3+ HCl → NH4Cl

c. 2O3 → 3O2

Explanation:

A redox reaction is a reaction in which there is loss or gain of electrons. As a result of that , there is a change in the oxidation number of the species involved in the reaction.

If we look at the species shown in the answer, there isn't any change in oxidation number as we move from left to right hence they are not redox reactions.

Redox reactions lead to change in oxidation number of species from left to right.

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Define electronegativity<br>

stira [4]

Answer: Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.

Explanation: Hope this helps!

6 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of pho

wlad13 [49]

<h3>Answer:</h3>

18.58 liters of hydrogen gas

<h3>Explanation:</h3>

We are given;

The equation;

3Mg + 2H₃(PO₄) → Mg₃(PO₄)₂ + 3H₂

Atoms of Magnesium = 7.179 x 10^23 atoms
Mass of phosphoric acid as 54.21 g

We are required to determine the volume of hydrogen gas produced;

Step 1; moles of Magnesium

1 mole of an element contains 6.02 × 10^23 atoms

therefore;

Moles of Mg = (7.179 x 10^23 ) ÷ (6.02 × 10^23)

= 1.193 moles

Step 2: Moles of phosphoric acid

moles = Mass ÷ Molar mass

Molar mass of phosphoric acid = 97.994 g/mol

Therefore;

Moles of Phosphoric acid = 54.21 g ÷ 97.994 g/mol

= 0.553 moles

Step 3: Determine the rate limiting reagent

From the mole ratio of Mg to Phosphoric acid (3 : 2);

1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

Therefore, phosphoric acid is the rate limiting reagent

step 4: Determine the moles of hydrogen produced

From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

Therefore; moles of Hydrogen = moles of phosphoric acid × 3/2

= 0.553 moles × 3/2

= 0.8295 moles

Step 5: Volume of hydrogen gas

1 mole of a gas occupies a volume of 22.4 liters at STP

Therefore;

Volume of Hydrogen = 0.8295 moles × 22.4 L/mol

= 18.58 Liters

Therefore; 18.58 liters of hydrogen gas will be produced

4 0

3 years ago

This graduated cylinder has numbers representing milliliters. A student says that the volume of the liquid inside is 21.8 millil

GaryK [48]

Explanation:

The answer for this question depends on the type of meniscus in the cylinder. If it is an upright meniscus like in water, the reading should be taken at the bottom of the meniscus. However if it is an inverted meniscus like in mercury, the reading should be taken at the top of the meniscus.

(Can you check and see if there's any pictures or information that is missing?)

3 0

3 years ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers

The lowest value of the henry's law for methane gas (ch4) will be obtained with __________ as the solvent and a temperature of _

NikAS [45]

The lowest value of the henry's law for methane gas (CH₄) will be obtained with H₂O as the solvent and a temperature of 349 K.

Henry's law: This law states that at a constant temperature, the amount of a gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas that in equilibrium with that liquid.

Mathematically it can be written as:

$C_{g}=k\times P_{g}$

So, for the methane gas , lowest value of the henry's law obtained at 349 K and with H₂O as the solvent.

3 0

3 years ago