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IRISSAK [1]
4 years ago
9

Which temperature represents the highest average kinetic energy of the particles in a sample of matter ?

Chemistry
1 answer:
Ymorist [56]4 years ago
6 0
Average kinetic energy is proportional to absolute temperature. This means that the higher the temperature the higher the kinetic energy.

This equation reflects that: KE avg = [3/2]kT, where k is the Boltzman constant.

So, you just have to checkg which of the given temperatures is the highest one..

Of course to compare you need to have them all in the same unit system.

Use absolute temperatures in Kelvin.

=> 27°C + 273.15 = 300.15 K.

Then, the highest temperature is 27°C and that represents the highest kinetic energy.

Answer is the option c. 27°C
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3 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
What is the molarity of 2.00 L of a solution that contains 14.6 g NaCl?
White raven [17]

Answer: The molarity of the solution is 0.125 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

moles of NaCl = \frac{\text {given mass}}{\text {Molar mass}}=\frac{14.6g}{58.5g/mol}=0.250mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.250mol}{2.00L}=0.125M

Therefore, the molarity of the solution is 0.125 M

8 0
3 years ago
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