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tino4ka555 [31]
3 years ago
10

Which statement describes a process or step in the rock cycle?

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
8 0
I think it’s letter b but not sure
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The fluid-filled area between the pleural layers is the.
Snowcat [4.5K]

The fluid filled area between the pleural layers is called the pleural cavity.

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3 years ago
Someone please answer the question attached.
nikdorinn [45]

Answer:

Option ( 1 )

Explanation:

This atom has a cub close pack structure, and therefore - the number of edge center present in X atoms =  4,

Number of unit cells present in X atoms = 6

____________________________________________________

Now the 6th coordination number of X atom = 6 * 4 = 24,

So respectively, the 3rd coordination number of X = 8 -

And thus the ratio between the 6th coordination number and the 3rd coordination number = 24 / 8 = 3,

Option ( 1 )

<u><em>Hope that helps!</em></u>

5 0
3 years ago
What is the percent composition of Ba(OH)2?
Alex17521 [72]
Element Symbol Atomic Mass
Barium Ba 137.327
Hydrogen H 1.00794
Oxygen O 15.9994


Hope this helps, good luck
3 0
3 years ago
Read 2 more answers
Can someone plz help me with these two questions?
bija089 [108]
For the first one, compound might mean more than one thing or i team. But mixture is a very specific definition for chemistry. And for the second one, it would be wrong because “pure” means only that substance, where as it would only be “orange juice”, or something else. But pure refers to only a clean not contaminated substance.
4 0
3 years ago
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
4 years ago
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