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kirill115 [55]
2 years ago
7

The element Osminum has a density of 22.6 g/cm3. the mass of a block of Osmium with three sides of 1.01cmx0.223 cm x 0.648 cm is

Chemistry
1 answer:
Alika [10]2 years ago
6 0
Volume of osmium = 1.01(0.223)(0.648) = 0.14595 cm3

Density = mass / volume
So density x volume = mass of osmium
22.6 x 0.14595 = 3.29845 g
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Explanation:

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Atoms=9.033*10^23 atoms

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How does a scientific theory differ from a guess or an opinion?​
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Decide whether the property measured was a chemical or physical property of X, if you can. If you don’t have enough information
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What will a balloon do to a cloth if its negatively charged
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2 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
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