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ArbitrLikvidat [17]

2 years ago

9

Which power source directs the electrons from oxidation-reduction reactions to flow through a device to give the device power? O

A. An electric motor B. A battery C. An electromagnet O D. An electric generator

1 answer:

Lady bird [3.3K]2 years ago

4 0

Answer:

B. A battery.

Explanation:

I just took the test

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What is the symbol for gold

Slav-nsk [51]

The elemental symbol of gold is denoted on the periodic table as being Au.

5 0

3 years ago

Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w

Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619= [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0

3 years ago

The following equilibrium is formed when copper and bromide ions are placed in a solution:

JulsSmile [24]

Answer:

A)

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

$heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)$

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

A)

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.

B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

Regards.

3 0

2 years ago

How a cold front affects weather.

miskamm [114]

Answer:

A cold front forms when a cold air mass pushes into a warmer air mass. Cold fronts can produce dramatic changes in the weather. ... As the cold front passes, winds become gusty. There is a sudden drop in temperature, and also heavy rain, sometimes with hail, thunder, and lightning.

4 0

3 years ago

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Non metal are usually poor conductor of hart and electricity . The are non -lustourous,non-sonours,non-malleable and are coloure

vekshin1

Check the solution in the attachment.

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2 years ago