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serg [7]
3 years ago
9

Two samples of carbon come into contact. A heat transfer will occur between sample A and sample B. What must be true for heat to

transfer from sample A to sample B?
The average kinetic energy of A is greater than that of B.
The average kinetic energy of B is greater than that of A.
The average kinetic energy of both samples is equal.
The average kinetic energy does not determine the direction of heat transfer.
Chemistry
2 answers:
jarptica [38.1K]3 years ago
8 0

Answer:

The average kinetic energy of A is greater than that of B.

Explanation:

The temperature of an object is directly proportional to the average kinetic energy of the particles in the object. For instance, for an ideal gas, we have

KE=\frac{3}{2}kT

where

KE is the kinetic energy

k is the Boltzmann constant

T the absolute temperature of the gas

Therefore, this means that in a hotter object the average kinetic energy of the particles is higher than the average kinetic energy of the particles in a colder object.

Moreover, the laws of thermodynamics tell us that heat is always transferred from a hotter object (higher temperature) to a colder object (lower temperature).

In this problem heat is transferred from sample A to sample B. Therefore, this means that object A has higher temperature, and therefore, higher average kinetic energy. So the correct answer is

The average kinetic energy of A is greater than that of B.

Alekssandra [29.7K]3 years ago
6 0

Answer:

A.The average kinetic energy of A is greater than that of B.

Explanation:

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Newton contributed to astronomy by:
SCORPION-xisa [38]

Answer:  D

Explanation:

This is the answer because everyone knows he discovered gravity and he conducted scientific experiments to prove them which he also used math for

Hope this helps

6 0
3 years ago
Read 2 more answers
What can be said about a reaction with H = 62.4 kJ/mol and S = 0.145 kJ/(mol·K)?
Zigmanuir [339]
Answer:

At 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

Explanation:

1) Variables:

G = Gibbs energy
H = enthalpy
S = entropy

2) Formula (definition)

G = H + TS

=> ΔG = ΔH - TΔS

3) conditions

ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction

4) Assuming the data given correspond to ΔH and ΔS

ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K

=>  T = [ΔH - ΔG] / ΔS

ΔG = 0 =>  T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K

This is, at 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

3 0
3 years ago
Read 2 more answers
PLS HELP ME I HAVE TILL 12
MariettaO [177]

Answer:

alright bud lets see hmm.... the answer is a. 90.5kpa

Explanation:

92.3 kPa - 1.82 kPa = 90.5 kPa

keep up your hope also corrected me if a am wrong in anyway! :)

3 0
3 years ago
A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at
lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

q = 1200 * 4.184 * 8.2 =  41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J  = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

8 0
2 years ago
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worty [1.4K]

Answer:

Ovary

Explanation:

ovaries produce the most estrogen in females.

5 0
2 years ago
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