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
The density of the rectangular block in g/mL is 7.0.
<u>Given the following data:</u>
- Mass of block = 22.8 gra1.94 kg
- Length of block = 3.21 cm
- Height of block = 1.84 in.
To find the density of the block in g/mL:
First of all, we would determine the volume of the rectangular block by using the following formula:
×
× 
<u>Conversion:</u>
1 in = 2.54 cm
5.83 in = X cm
Cross-multiplying, we have:

×
× 
Volume = 277.16 cubic centimeters.
<u>Note</u>: Milliliter (mL) is the same as cubic centimeters.
1000 grams = 1 kg
Y grams = 1.94 kg
Cross-multiplying, we have:
Y = 1940 grams
Now, we can find the density:

<em>Density </em><em>= 7</em><em>.0 g/mL</em>
Therefore, the density of the rectangular block in g/mL is 7.0.
Read more: brainly.com/question/18320053
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
A)3 bcuz paper filters are used to separate solids from liquids
b) 2 bcuz it is a process of separating a pure liquid from a mixture of other liquids.And it works when it has different boiling point
Carbon atoms are extremely small and are one of the only atoms that are structurally stable enough to form various different kinds of macromolecules.