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PtichkaEL [24]
3 years ago
15

Determine if the following statements are true or false. True Standard reduction potential is an intensive property True Reducti

on takes place at the anode True The half reaction with the lower standard reduction potential will be at the cathode in a galvanic cell True The half reaction with the higher standard reduction potential will be at the cathode in a galvanic cell
Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

Standard reduction potential is an intensive property---- True

Reduction takes place at the anode ----- False

The half reaction with the lower standard reduction potential will be at the cathode in a galvanic cell ------false

The half reaction with the higher standard reduction potential will be at the cathode in a galvanic cell  ------  True

Explanation:

An intensive property is a property of a substance which is inherent in it and part of its nature. It does not depend on the amount of substance present in the substance. Standard reduction potential is an intensive property.

In a galvanic cell, oxidation takes place at the anode and reduction takes place at the cathode. At the anode, the electrode potential is more negative (an oxidation) while at the cathode the reduction potential is less negative (a reduction).

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(a)Determine the number of KNO3 molecules in 0.750 mol KNO3.
svp [43]

A. The number of molecules in 0.750 mole of KNO₃ is 4.515×10²³ molecules

B. The mass (in milligrams) of 2.39×10²⁰ molecules of Ag₂SO₄ is 124 mg

C. The number of molecules in 3.429 g of NaHCO₂ is 3.04×10²² molecules

<h3>Avogadro's hypothesis </h3>

1 mole of substance = 6.02×10²³ molecules

<h3>A. How to determine the number of molecules </h3>

1 mole of KNO₃ = 6.02×10²³ molecules

Therefore,

0.750 mole of KNO₃ = 0.75 × 6.02×10²³

0.750 mole of KNO₃ = 4.515×10²³ molecules

<h3>B. How to determine the mass of Ag₂SO₄</h3>

6.02×10²³ molecules = 312 g of Ag₂SO₄

Therefore,

2.39×10²⁰ molecules = (2.39×10²⁰ × 312) / 6.02×10²³

2.39×10²⁰ molecules = 0.124 g

Multiply by 1000 to express in mg

2.39×10²⁰ molecules = 0.124 g × 1000

2.39×10²⁰ molecules = 124 mg of Ag₂SO₄

<h3>C. How to determine the number of molecules </h3>

68 g of NaHCO₂ = 6.02×10²³ molecules

Therefore,

3.429 g of NaHCO₂ = (3.429 × 6.02×10²³) / 68

3.429 g of NaHCO₂ = 3.04×10²² molecules

Learn more about Avogadro's number:

brainly.com/question/26141731

5 0
2 years ago
Determine the value of x and write the complete chemical formula of the hydrate
Annette [7]
In order to determine the formula of the hydrate, [Anhydrous Solid⋅xH2O], the number of moles of water per mole of anhydrous solid (x) will be calculated by dividing the number of moles of water by the number of moles of the anhydrous solid (Equation 2.12.
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3 years ago
How many grams of sodium chromate, Na2CrO4, are needed to react completely with
JulsSmile [24]

Answer:

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

Explanation:

Step 1: data given

Mass of silver nitrate AgNO3 = 56.7 grams

Molar mass AgNO3 = 169.87 g/mol

Molar mass of Na2CrO4 = 161.97 g/mol

Step 2: The balanced equation

2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 56.7 grams / 169.87 g/mol

Moles AgNO3 = 0.334 moles

Step 4: Calculate moles Na2CrO4 moles needed

For 2 moles AgNO3 we need 1 mol Na2CrO4 to produce 1 mol Ag2CrO4 and 2 moles NaNO3

For 0.334 moles AgNO3 we need 0.334 / 2 = 0.167 moles Na2CrO4

Step 5: Calculate mass Na2CrO4

Mass Na2CrO4 = 0.167 moles * 161.97 g/mol

Mass Na2CrO4 = 27.0 grams

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

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Answer:

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