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Artemon [7]
2 years ago
13

Which statement identifies the function of a cathode in every in every electrochemical cell

Chemistry
1 answer:
torisob [31]2 years ago
6 0

The answer is B because <span>It would be useful to memorize that sentence. Once you know that, you can figure out whatever else happens at the anode, the cathode, in the solution, and in the external circuit.</span>
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How many grams of Mg(NO3)2 would be produced? please show how you got the answer​
Len [333]

Answer:

148 grams of relative atomic mass

Explanation:

magnesium atomic mass : 24

nitrogen : 14

oxygen : 16

24 × 1

14 × 1 × 2

16 × 3 × 2

24 + 28 + 80 = 148 grams

7 0
3 years ago
What is Charles’s law? State the definition of the law in words. What are the assumptions of Charles’s law? Write mathematical e
Leviafan [203]

Answer:

The volume of a given mas of a gas is directly proportional to the temperature if the pressure remains constant

V is directly proportional to T

V=1/T

V=constant/T

Explanation:

4 0
3 years ago
Can someone answer this please?
krok68 [10]
What is the question?? 
3 0
3 years ago
The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
nadezda [96]

Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

7 0
3 years ago
When C3H8 burns completely in an excess of oxygen , the products formed are?
Marizza181 [45]
Water and carbon dioxide respectively

6 0
3 years ago
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