It is A) 1,482 cm3 ..............
The normality that would be calculated will be to high because the change in volume will be greater than the actual change in volume. if the buret tip is not filled when reading the initial volume, the actual volume should be lesser with that reading. so if you will you the higher reading the change in volume or the volume you use in titration will be higher than the actual
When given percents for an empirical formula problem, first consider all the percents as grams.
58.82% carbon —> 58.82 g carbon
27.45% nitrogen —> 27.45 g nitrogen
13.73% hydrogen —> 13.73 g hydrogen
Then convert the grams of all the elements to moles, based on their molar masses.
Carbon - 58.82 g / 12.01 g/mol = 4.898 mol carbon
Nitrogen - 27.45 g / 14.01 g/mol = 1.959 mol nitrogen
Hydrogen - 13.73 g / 1.008 g/mol = 13.62 mol hydrogen
Then divide all of the mole numbers by the smallest number of moles, which is in this case, the 1.959 mol of nitrogen.
Carbon - 4.898 / 1.959 = 2.5
Nitrogen - 1.959 / 1.959 = 1
Hydrogen - 13.62 / 1.959 = 7
You want whole numbers for all of your mole numbers, so multiply all of them by 2, since 2.5 isn’t a whole number.
Final answer: C5N2H14
Three double bonds and four single bonds, or 3(+)+4=7+3, are present in cumulene.
Sigma bond between the hydrogen and carbon on the end-side of the cumulene is visible. Carbon and hydrogen are linked by a total of four sigma bonds. Carbon is joined to carbon in the center by 3 sigma bonds and 3 pi bonds. Cumulene now has a total of 7 sigma bonds and 3 pi bonds. A cumulene is a molecule in organic chemistry that has three or more cumulative (consecutive) double bonds. sp2 hybrid orbital of C overlapping with 2p orbital of O and one bond make up the bond.
In the case of compounds with two bonds, one bond is a sigma bond and the other is a pi bond. A triple bond indicates that the chemicals.
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This is an incomplete question, here is a complete question.
Calculate the volume in milliliters of a 1.29 mol/L iron(II) bromide solution that contains 275 mmol of iron(II) bromide . Round your answer to significant 3 digits.
Answer : The volume of iron(II) bromide solution is, 
Explanation : Given,
Concentration of iron(II) bromide = 1.29 mo/L
Moles of iron(II) bromide = 275 mmol = 0.275 mol
conversion used : 1 mmol = 0.001 mol
Now we have to calculate the volume of iron(II) bromide.
Now put all the given values in this formula, we get:
Thus, the volume of iron(II) bromide solution is,
