Answer:
the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula
Explanation:
Mass of CO2 obtained = 3.14 g
Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol
The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g
Mass of H2O obtained = 1.29 g
Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol
The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g
% by mass of carbon = 0.857/1 ×100 = 85.7 %
% by mass of hydrogen = 0.0717/1 × 100 = 7.17%
Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %
Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.
Answer:
See the explanation
Explanation:
In this case, in order to get an <u>elimination reaction</u> we need to have a <u>strong base</u>. In this case, the base is the phenoxide ion produced the phenol (see figure 1).
Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).
Finally, the phenoxide will attack the <u>primary carbon</u> attached to the Cl. The C-Cl bond would be broken and the C-O would be produced <u>at the same time</u> to get a substitution (see figure 1).
Answer: 72.93 litres
Explanation:
Given that:
Volume of gas (V) = ?
Temperature (T) = 24.0°C
Convert 24.0°C to Kelvin by adding 273
(24.0°C + 273 = 297K)
Pressure (P) = 1.003 atm
Number of moles (n) = 3 moles
Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1
Then, apply ideal gas equation
pV = nRT
1.003 atm x V = 3.00 moles x 0.0821 atm L K-1 mol-1 x 297K
1.003 atm•V = 73.15 atm•L
Divide both sides by 1.003 atm
1.003 atm•V/1.003 atm = 73.15 atm•L/1.003 atm
V = 72.93 L
Thus, the volume of the gas is 72.93 litres
Your answer is Canada for sure I hope you make a 100 :))
Answer:
the equation is unbalanced
