Answer:
Given, 0.29 g of hydrocarbon produces 448ml of CO2 at STP. then, C2H5 is the emperical formula of hydrocarbon . n = 2 , hence, molecular formula will be C4H10
Answer:
2.893 x 10⁻³ mol NaOH
[HCOOH] = 0.5786 mol/L
Explanation:
The balanced reaction equation is:
HCOOH + NaOH ⇒ NaHCOO + H₂O
At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.
The moles of base added is calculated as follows:
n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH
Extra significant figures are kept to avoid round-off errors.
Now we relate the amount of NaOH to the amount of HCOOH through the molar ratio of 1:1.
(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH
The concentration of HCOOH to the correct number of significant figures is then calculated as follows:
C = n/V = (2.892992 mmol) / (5.00 mL) = 0.5786 mol/L
The question also asks to calculate the moles of base, so we convert millimoles to moles:
(2.892992 mmol NaOH)(1 mol/1000 mmol) = 2.893 x 10⁻³ mol NaOH
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Answer:
96%
Explanation:
Step 1: Write the balanced neutralization reaction
Al(OH)₃ + 3 HCl ⇒ AlCl₃ + 3 H₂O
Step 2: Calculate the theoretical yield of AlCl₃
According to the balanced equation, the mass ratio of Al(OH)₃ to AlCl₃ is 81.03:133.34.
28 g Al(OH)₃ × 133.34 g AlCl₃/81.03 g Al(OH)₃ = 46 g AlCl₃
Step 3: Calculate the percent yield of AlCl₃
The real yield of AlCl₃ is 44 g. We can calculate the percent yield using the following expression.
%yield = real yield / theoretical yield × 100%
%yield = 44 g / 46 g × 100% = 96%
