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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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Calculate the mole fraction of each component in a solution with 6.87 g of sodium chloride (NaCl) dissolved in 65.2 g of water.
zheka24 [161]

Answer:

mole fraction of NaCl = 0.03145.

mole fraction of water = 0.9686.

Explanation:

  • Mole fraction is an expression of the concentration of a solution or mixture.
  • It is equal to the moles of one component divided by the total moles in the solution or mixture.
  • The summation of mole fraction of all mixture components = 1.

mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles).

<em>no. of moles of NaCl = mass/molar mass </em>= (6.87 g)/(58.44 g/mol) = 0.1176 mol.

<em>no. of moles of water = mass/molar mass</em> = (65.2 g)/(18.0 g/mol) = <em>3.622 mol.</em>

<em></em>

∴ mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles) = (0.1176 mol)/(0.1176 mol + 3.622 mol) = 0.03145.

<em>∵ mole fraction of NaCl + mole fraction of water = 1.0.</em>

∴ mole fraction of water = 1.0 - mole fraction of NaCl = 1.0 - 0.03145 = 0.9686.

7 0
3 years ago
Be able to describe the relationship between Freedom of Movement and phase change.
lisabon 2012 [21]

Ion knoe Wdym by “be able to describ’ so ima put it in my own words idr lol:)

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I can explain how transferring kinetic energy in and out of a substance can cause a change

7 0
3 years ago
How to answer all these questions?
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You have to do it to see the results
8 0
3 years ago
The recommended daily allowance (rda of calcium is 1.2 g. calcium carbonate contains 12.0% calcium by mass. how many grams of ca
defon
1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!

2) Mass fraction of this is excessive data.

3) The solution is:

m(Ca)=1.2 g

m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)

m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
4 0
3 years ago
Read 2 more answers
Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere
Colt1911 [192]
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
            = (2x10^-16)/(1x10^-7)^2
             = 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4 
when POH = -㏒[OH-]
            4  = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
          = 2x10^-16 / (1x10^-4)^2
          = 2x10^-8 M
c) at PH= 14 
when POH = 14-PH
                   = 14 - 14 
                   = 0
when POH = -㏒[OH]
              0 = - ㏒[OH]
∴[OH] = 1 m 
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
          = (2x10^-16) / 1^2
          = 2x10^-16 M


8 0
3 years ago
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