It is hard to answer this because not much information is given.
If we consider a combustion reaction of Methane:
The balanced equation is:
CH4 + 2O2 ---> 2H2O + CO2
The rate of appearance of H2O is rH2O, rate of disappearance of O2 is -rO2
(rH2O)^2 = (-rO2)^2
rH2O = -rO2
Answer:Hope this helps!
Explanation:
You can use a flame test to help identify the composition of a sample. The test is used to identify metal ions (and certain other ions) based on the characteristic emission spectrum of the elements. The test is performed by dipping a wire or wooden splint into a sample solution or coating it with the powdered metal salt. The color of a gas flame is observed as the sample is heated. If a wooden splint is used, it's necessary to wave the sample through the flame to avoid setting the wood on fire. The color of the flame is compared against the flame colors known to be associated with the metals.
Answer:
Mass = 0.697 g
Explanation:
Given data:
Volume of hydrogen = 1.36 L
Mass of ammonia produced = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
First of all we will calculate the number of moles of hydrogen:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K
1.36 atm.L = n × 22.43 atm.L/mol
n = 1.36 atm.L / 22.43 atm.L/mol
n = 0.061 mol
Now we will compare the moles of hydrogen and ammonia:
H₂ : NH₃
3 : 2
0.061 : 2/3×0.061 = 0.041
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.041 mol × 17 g/mol
Mass = 0.697 g
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).
