Answer: Thomson
Explanation: It verified J. J. Thomson's work on the atomic structure.
2H2(g) + O2(g) → 2H2O(1) 0 260 g 0.2068 0.180 g 2008
When 45.0 g of CH4 reacts with excess O2, the actual yield of CO2 is 118 g. What is the percent yield? CHA(g) + 2O2(g) - CO2(g) + 2H2O(g) 73.6% 67.9% 95.2% 86.4%
For the reaction: 2503(g) + 790 kcal - 25(s) + 3O2(g), how many kcal are needed to form 1.5 moles O2(g)? 790 kcal 395 kcal 2370 kcal 411 kcal
When 3 moles of Ny are mixed with 5 moles of H2 the limiting reactant is N2(g) + 3H2(g) - 2NH3(g) H2 NH3 ОООО H20 O N₂
Given:
P1 = 13.0 atm
T1 = 20 °C
T2 = 102 °C
Required:
P2 of oxygen
Solution:
At constant volume,
we can apply Gay-Lussac’s law of pressure and temperature relationship
P1/T1=P2/T2
(13.0 atm) / (20 °C)
= P2 / (102 °C)
P2 = 66.3 atm
The answer is not in the choices given.
A- Identify the mixture:
The mixture of powdered charcoal and powdered sugar is considered as a homogeneous mixture. This means that you cannot identify the components with naked eye as they are uniformly distributed in the mixture.
B- Separate components:
You ca separate the charcoal powder from the sugar powder using the following steps:
1- add water. Sugar will dissolve in water while charcoal won't.
2- filter the solution where the powdered charcoal will remain on the filter paper and the solution of powder will pass through.
3- boil the sugar solution (above 100 degrees celcius). The water will evaporate and the sugar will precipitate.
The answer is 6 moles of water will be produced.
