Multiply .800 moles of O2 by Avagadro's number divided by 1 mole. This will get rid of the moles on the bottom and leave you with molecules. So technically .800 times 6.02x10^23.
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.
Answer:
1. B, D,
2.A, F
Explanation:
1. According to the law of conservation of mass, In a course of chemical reaction, matter can neither be created nor destroyed but can be changed from one form to another. This means the amount of matter at the begining and ending of a reaction must be thesame.
2. Chemical reaction is not easily reversible. when gas is produced, provided the reaction system is an open system, the gas cannot be recovered and the reactants cannot be recovered from the products. likewise color change are attributed to chemical reaction
Answer:
Mass of solid = 189.141 gram
Explanation:
Given:
Total volume = 93 ml
Mass of liquid = 33.7 gram
Density of liquid = 0.865 g/ml
Density of solid = 3.50 g/ml
Find:
Mass of solid = ?
Computation:
Volume of liquid = Mass of liquid / Density of liquid
Volume of liquid = 33.7 / 0.865
Volume of liquid = 38.9595 ml
Volume of solid = Total volume - Volume of liquid
Volume of solid = 93 - 38.9595
Volume of solid = 54.0405 ml
Mass of solid = Volume of solid × Density of solid
Mass of solid = 54.0405 ml × 3.50 g/ml
Mass of solid = 189.141 gram
Answer:
M
Explanation:
Concentration of ![Cu^{2+} = [Cu(NO_3)_2]](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%20%3D%20%5BCu%28NO_3%29_2%5D)
= 0.020 M
Constructing an ICE table;we have:
![Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%2B4NH_3_%7Baq%7D%20%5Crightleftharpoons%20%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D_%7B%28aq%29%7D)
Initial (M) 0.020 0.40 0
Change (M) - x - 4 x x
Equilibrium (M) 0.020 -x 0.40 - 4 x x
Given that: 
![K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}](https://tex.z-dn.net/?f=K_f%20%7D%20%3D%20%5Cfrac%7B%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D%7D%7B%5BCu%5E%7B2%2B%7D%5D%5BNH_3%5D%5E4%7D)
Since x is so small; 0.40 -4x = 0.40
Then:
M
