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ZanzabumX [31]
3 years ago
8

If you burn 52.0 g of hydrogen and produce 465 g of water, how much oxygen reacted?

Chemistry
1 answer:
Nadusha1986 [10]3 years ago
7 0
Well, since matter doesn't simply poof into existence and water is H2O, 465g of water is a combination of 52.0g of H and 413g O
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Which of the following pairs of elements could possibly be in the same group? X has a 1+ ion; Y has a 1- ion. X tends to form a
exis [7]
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
5 0
3 years ago
Read 2 more answers
___ Au₂S₃ + ___ H₂ → ___ Au + ___ H₂S
Natalija [7]

Answer:

2Au₂S₃ +  6H₂ → 4Au + 6H₂S

Explanation:

Balancing:

2Au₂S₃ +  6H₂ → 4Au + 6H₂S

6 0
3 years ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0
3 years ago
How are carbohydrates and lipids similar? how are they different?
Ludmilka [50]
"Lipids<span> are like </span>carbohydrates<span> in way that the true fats contain only carbon, hydrogen, and oxygen. Both </span>carbohydrates and lipids<span> act as the main fuels and energy storage compounds of the human body. They are also called SACCHARIDES and grouped as: Monosaccharides, Disaccharides, Trisaccharides, Polysaccharides."

Source credit: </span>https://www.quora.com/What-are-the-differences-and-similarities-between-carbohydrates-and-lipids
6 0
2 years ago
What is the volume of 68.0 g of ether if the density of ether is 0.72 g/mL?
umka2103 [35]

Answer:

94.44

Explanation:

Volume is equal to Mass/Density so therefore, you do the mass which is 68.0 g/0.72 g/mL which is the density and get 94.44 mL because the g cancel each other out when it comes to the label!

3 0
2 years ago
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