Answer:
Close to the calculated endpoint of a titration - <u>Partially open</u>
At the beginning of a titration - <u>Completely open</u>
Filling the buret with titrant - <u>Completely closed</u>
Conditioning the buret with the titrant - <u>Completely closed</u>
Explanation:
'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.
As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.
Each isotope has a unique rate of decay, making them suitable for determining the dates of ancient artifacts. The answer is "rate of decay of the isotope."
Yes, it's important that acids and bases have different properties. Because they'll react differently.
There are three major steps for finding an empirical formula from a combustion reaction.
1.) Determine the grams of carbon and hydrogen from the given data.
C: 23.98 g x (12.011 g / 44.01 g) = 6.54 g C
H: 4.91 g x (2.0158 g / 18.02 g) = 0.55 g H
Determine the grams of oxygen in the sample by subtracting the mass of the compound given from the total mass solved earlier.
O: 10.0 - (6.54 + 0.55) = 2.91 g O
2) Convert the values in step 1 to moles.
C: 6.54 g / 12.01 g / mol = 0.54 mol
H: 0.55 g / 1.01 g/mol = 0.54 mol
O: 2.91 g / 16.00 g/mol = 0.18 mol
3) Divide each by the lowest value calculated in step 2
C: 0.54 mol / 0.18 mol = 3
H: 0.54 mol / 0.18 mol = 3
O: 0.18 mol / 0.18 mol = 1
Thus, the empirical formula is C3H3O.
Answer:
CH4 + 2O2 -> CO2 + 2H2O
Explanation:
This reaction involves the combustion of methane (CH4)
General form of balancing reactions involving combustion of hydrocarbons is
CxHy + [x+(y/4)]O2 -> xCO2 + (y/2)H2O
x= the number of carbon atoms in the compound = 1
y= the number of hydrogen atoms in the compound = 4
CH4 + [(1+(4/4)]O2 -> CO2 + (4/2)H20
CH4 + (1+1)O2 -> CO2 + 2H2O
CH4 + 2O2 -> CO2 + 2H2O
