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weeeeeb [17]
3 years ago
9

Why is carbon tetrachloride a liquid at room temperature

Chemistry
1 answer:
miv72 [106K]3 years ago
3 0
Because I'm the warm it melts in the cold it's frozen
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1. Convert each of the following Celsius temperatures to Kelvin. a) 27°C b) 100°C c) 0°C​
valkas [14]

Answer:

a) 300K

b) 373K

c) 273K

Explanation:

to go from °C to K all you have to do is add 273.

5 0
2 years ago
A gas with a volume of 250 mL at a temperature of 293K is heated to 324K. What is the new volume of the gas?
Natalija [7]

Answer:

The new volume of the gas is 276.45 mL.

Explanation:

Charles's law indicates that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases, and as the temperature decreases, the volume of the gas decreases.

Charles's law is a law that mathematically says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

\frac{V}{T} =k

Analyzing an initial state 1 and a final state 2, it is satisfied:

\frac{V1}{T1} =\frac{V2}{T2}

In this case:

  • V1= 250 mL
  • T1= 293 K
  • V2= ?
  • T2= 324 K

Replacing:

\frac{250 mL}{293 K} =\frac{V2}{324 K}

Solving:

V2=324 K*\frac{250 mL}{293 K}

V2= 276.45 mL

<em><u>The new volume of the gas is 276.45 mL.</u></em>

4 0
3 years ago
What is the molar mass of a substance?
OlgaM077 [116]

oops pls forgive me I accidentally did the wrong question.

4 0
2 years ago
When the pH value of a solution is changed from 2
inna [77]

increases my factor of 10

6 0
3 years ago
Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati
Korvikt [17]

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{vap}}{T_b}

where,

\Delta S = change in entropy

\Delta H_{vap} = change in enthalpy of vaporization = 40.5 kJ/mol

T_b = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{vap}}{T_b}

\Delta S=\frac{40.5kJ/mol}{352K}

\Delta S=115J/mol.K

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

8 0
3 years ago
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