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3 years ago

The combustion of Methane produces energy. Where does this energy come from

Chemistry

2 answers:

hram777 [196]3 years ago

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Firdavs [7]3 years ago

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A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume

jenyasd209 [6]

Volume of Argon V1 = 5.0 L
Pressure of Argon P1 = 2 atm
Final temperature T2 = 30 C = 30 + 273 = 303 K
Volume at final temperature V2= 6 L
Pressure at final temperature P2 = 8 atm
We know that (P1 x V1) / T1 = (P2 x V2) / T2
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48
Initial Temperature T1 = 3030 / 48 = 63.12
Initial Temperature = -209. 8 C

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2 years ago

In terms of their electron configurations, why is cesium more likely to lose its valence electron than potassium?

Harlamova29_29 [7]

Explanation:

use the term electron sheilding, the more electrons between the valence el3ctron and nucleus the easier to lose the valence electron (more sheilding = easier to lose)

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3 years ago

HELPPPPP PLEASEEE ILL MARK YOU BRAINLIESTTTTTT!!!!!!!!!!!!!!!!!!!!!!!!!!

azamat

Answer:

Ionic bonding occurs when atoms either gain or lose one or more valence electrons, resulting in the atom having either a negative or positive charge.

Through ionic bonding, an atom of each element will combine with the other to form a molecule, which is more stable since it now has a zero charge.

Explanation:

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3 years ago

Read 2 more answers

What is the process of breaking down food to release energy by adding oxygen

a_sh-v [17]

I believe this process is called cellular respiration.

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3 years ago

Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for

eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

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3 years ago