2 Li(s) +Cl₂→ 2 Li⁺ (aq) + 2Cl⁻ (aq)
The cell potential of the reaction above is +4.40V
<em><u>calculation</u></em>
Cell potential =∈° red - ∈° oxidation
in reaction above Li is oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04
Cl is reduce from oxidation state 0 to -1 therefore the ∈°red = +1.36 V
cell potential is therefore = +1.36 v -- 3.04 = + 4.40 V
Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.
Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.

Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:

Therefore, density of metal is 19.33 g/mL.
The number of moles in 3.612 x 10²⁴ molecules of CaO is 6 moles.
<h3>
Number of moles in the molecules</h3>
The number of moles in 3.612 x 10²⁴ molecules of CaO is calculated as follows;
6.02 x 10²³ molecules = 1 mole
3.612 x 10²⁴ molecules = ?
= (3.612 x 10²⁴ ) / (6.02 x 10²³ )
= 6 moles
Thus, the number of moles in 3.612 x 10²⁴ molecules of CaO is 6 moles.
Learn more about number of moles here: brainly.com/question/15356425
First choice: PhH4 has a lower boiling point
Second: SiF4 higher
Third: CBr4 higher
Answer:
option B.
Explanation:
Given,
V₁ = 156 L
P₁ =2 atm
Now, in the cylinder
P₂ = ?
V₂ = 36
Using relation between pressure and volume



Hence, pressure is equal to 8.67 atm.
Hence, the correct answer is option B.