Mass/volume = density
mass = (440 mg)*(1g)/(1000mg) = 0.440g
volume = (1000cm)(1000cm)(t)
where t = thickness
density = 2.70 g/cm^3 = (0.440g)/((1000cm)(1000cm)(t))
multiply both sides by 't' and divide both sides by (2.70g/cm^3)
t = (0.440) / ((1000cm)(1000cm)(2.70)) = 1.629x10^-7 cm
t = (1.629 x 10^-7 cm)*(1000000 micrometers)/(1 cm) = 0.1629 micrometers
Answer is t = 0.1629 micrometers
Answer:
O
Explanation:
The atoms lose energy during a change of state, but can still slide past each other; gas to a liquid.
Answer: I am confident the answer is B
Explanation:
forgive me if im wrong
Answer:
Note that melting and vaporization are endothermic processes in that they absorb or require energy, while freezing and condensation are exothermic process as they release energy.
Answer:
The answer to the question is
The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits
Explanation:
To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure
At the first trial the mixture contains
3.6 atm CO
1.2 atm H₂O (g)
Total pressure = 3.6+1.2= 4.8 atm
which gives
3.36 atm CO
0.96 atm H₂O (g)
0.24 atm H₂ (g)
That is
CO+H₂O→CO(g)+H₂ (g)
therefore the mixture contained
0.24 atm CO₂ and the total pressure =
3.36+0.96+0.24+0.24 = 4.8 atm
when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂
At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857
adding 1.8 atm CO gives 4.46 atm hence we have
(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857
which gives x = 0.031 atm or x = -0.6183 atm
Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
