Answer:
Mass of chemical = 1.5 mg
Explanation:
Step 1: First calculate the concentration of the stock solution required to make the final solution.
Using C1V1 = C2V2
C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution
C1 = C2V2/V1
C1 = (6 * 25)/ 0.1
C1 = 1500 ng/μL = 1.5 μg/μL
Step 2: Mass of chemical added:
Mass of sample = concentration * volume
Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL
Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg
Therefore, mass of sample = 1.5 mg
First you must write a balanced chemical equation.
C3H8 + 5O2 --> 3CO2 + 4H2O
From there, we can set up the stoichiometry equation to solve.
g O2= 70.2 g C3H8 X (1 mol C3H8/44.0962g C3H8) X (5 mol O2/1 mol C3H8) X (31.998g O2/1 mol O2)
Now solve, and you should get 254.7 g O2. Hope this helped!
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.
The
balanced chemical reaction will be:
C2H2 + 3/2O2 = 2CO2 +H2O
We are given the amount of C2H2 to be burned. This will be
our starting point.
<span>35.0 mol C2H2 ( 2 mol CO2/1 mol C2H2) = 70.0 mol O2</span>
<span>Therefore, the CO2 produced is 70.0 moles.</span>
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