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yKpoI14uk [10]
2 years ago
14

How much heat is lost when 1277 g of H20 at 400 K is changed into ice at 263 K? Plz help I’ll give you brainliest 20 points

Chemistry
1 answer:
OLga [1]2 years ago
6 0

The heat lost is -7.32*10^-^5J

The heat lost when the ice is cooled from 400k to 263K can be calculated using the formula of heat transfer.

<h3>Heat Transfer</h3>

This is the heat transferred from a body of higher temperature to a body of lower temperature.

Q = mc(\delta)T

  • Q = Heat Transfer
  • m = mass = 1277g
  • ΔT  = change in temperature

\delta T = (400 - 273) - (263 - 273) = \\&#10;\delta T = T_2 - T_1\\&#10;\delta T = -10 - 127\\&#10;\delta T = -137^0C

We converted the temperature from kelvin scale into Celsius scale and find the change in temperature.

Solving for heat transfer

Q = mc\delta T\\&#10;Q = 1277 * 4.186 * -137\\&#10;Q = -732336.514J

The heat loss is approximately -7.32*10^-^5J

Learn more on heat transfer here;

brainly.com/question/16055406

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Answer:

The value of the equilibrium constant KC is 1.244

Explanation:

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C

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Step 1: The balanced equation

A+2B<->C

Step 2: The initial concentrations

[A] = 0.550 M

[B]= 1.40 M

[C] = 0.600 M

Step 3: The concentraions at equilibrium

[A] = 0.550 -X = 0.430 M

[B]= 1.40 -2X M

[C] = 0.600 + X = 0.720 M

X = 0.120 M

[A] = 0.550 - 0.120 = 0.430 M

[B]= 1.40 -2*0.120 =  1.16 M

[C] = 0.600 + 0.120 = 0.720 M

Step 4: Calculate Kc

Kc = [C] / [A][B]²

Kc = 0.720 / (0.430*1.16²)

Kc = 1.244

The value of the equilibrium constant KC is 1.244

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Explanation:

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Pls help Would be much appreciated:)
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Answer:

Ok so,  b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...

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iii. Calculate the standard potential (voltage) of the cell

Look up the reduction potential,

E

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red

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