How you would prepare 200.0mL of a 0.25M NaOH solution. a) 2g b) 4g c) 8g d) 6g
1 answer:
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Answer:
1-Ethyl-3-methylidenecyclopentane
Step-by-step explanation:
Formula = C₈H₁₄. An alkane has formula C₈H₁₈. ∴ X contains 2 double bonds, 2 rings, or 1 ring and 1 double bond.
X absorbs only 1 mol of hydrogen. ∴ X contains 1 ring and 1 double bond.
Hydrogenation gives 1-ethyl-3-methylcyclopentane.
Ozonolysis gives formaldehyde, so X must contain a =CH₂ group.
Hydrogenation of X converted the =CH₂ to -CH₃.
X is 1-ethyl-3-methylidenecyclopentane.
You can see the reactions in the image below.
This is an incomplete question, here is a complete question.
A quantity of CO gas occupies a volume of 0.48 L at 1.6 atm and 282 K. The pressure of the gas is lowered and its temperature is raised until its volume is 2.1 L. Find the density of the CO under the new conditions?
Answer : The density of the CO under the new conditions is, 0.213 g/L
Explanation :
First we have to determine the moles of CO gas by using ideal gas equation.
where,
P = pressure of gas = 0.90 atm
V = volume of gas = 0.64 L
T = temperature of gas = 3 07 K
R = gas constant = 0.0821 L.atm/mol.K
n = number of moles of gas = ?
Now put all the given values in the above formula, we get:
Now we have to calculate the mass of CO gas.
Molar mass of = 28g/mole
Now we have to calculate the density of CO gas.
Therefore, the density of the CO under the new conditions is, 0.213 g/L