Answer:
Condenses at 27.25K.
Freezes at 24.65K.
Explanation:
In order to solve this above question, there is is need to make use of the following equation. The main idea here is to convert degree celsius to Kelvin. Hence,
0°C + 273.15 = 273.15K---------------------(1).
Therefore, we will make use of the above equation (1) and slot in the values for at degree celsius at which it condenses and at degree celsius at which it freezes.
So, we have at temperature at which it condenses:
-245.9°C + 273.15 = 27.25K.
Also, we have at temperature at which it freezes.
-248.5°C + 273.15 = 24.65K.
Although coal and diamonds are both composed of carbons alone, they differ in their properties because of the bonds present in each substance. Diamonds are composed of covalent bonds which are strong and is crystalline in nature which makes it sturdier. Coal is amorphous which allows the slipping of irregular layers to each other making coal brittle
Complete Question
The complete question is shown on the first uploaded image
Answer:
The pressure is 
Explanation:
From the question we are told that
The mass of the carbon monoxide is 
The temperature at which takes place 
The volume of the sealed vessel is 
Generally the ideal gas law is mathematically represented as

Where R is the gas constant with value 
n is the number of moles of carbon monoxide which is mathematically evaluated as

where
is the molar mass of carbon monoxide which is a constant with value

So 

Now Making P the subject we have


Equation: M1V1 = M2V2
Where M = concentration & V = volume
Step 1: Write down what is given and what you are trying to find
Given: M1 = 6.00M, V1 = 2.49mL, and V2 = 50.0mL
Find: M2
Step 2: Plug in the values into the equation
M1V1 = M2V2
(6.00M)(2.49mL) = (M2)(50.0mL)
Step 3: Isolate the variable (Divide both sides by 50.0mL so M2 is by itself)
(6.00M)(2.49mL) / (50.0mL) = M2
Answer: M2 = 0.30M
*Don't forget sig figs & units!
Answer:C
Explanation:
It would be the most visually understandable.It would be easyer to understand what you are modeling.