<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.
<u>Explanation:</u>
We are given:
The substance having highest positive 
potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.
Aluminium will undergo oxidation reaction and will get oxidized.
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the 
of the reaction, we use the equation:
Hence, the standard electrode potential of the cell is 4.53 V.
Answer:
29 L.
Explanation:
Hello!
In this case, considering that we are performing a conversion by which the time should be cancelled out to obtain liters, we first need to convert the seconds on bottom to hours and then the volume on top to liters, just a shown down below:
Which turns out 29 L with 2 significant figures.
Best regards!
Answer:
d= 14.007 amu
Explanation:
Abundance of N¹⁴ = 99.63%
Abundance of N¹⁵ = 0.37%
Atomic mass of N¹⁴ = 14.003 amu
Atomic mass of N¹⁵ = 15.000 amu
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (14.003 × 99.63)+(15.000× 0.37) /100
Average atomic mass = 1395.12 + 5.55 / 100
Average atomic mass = 1400.67/ 100
Average atomic mass = 14.007 amu.
I'm guessing you mean Cr (Chromium) in the product instead of Ct.
<span>Pb(NO3)2 + Na2(CrO4) = Pb(CtO4) + 2Na(NO3)</span>
Answer:
2100 kPa
Explanation:
The temperature is constant, so the only variables are pressure and volume.
We can use Boyle’s Law.
p₁V₁ = p₂V₂ Divide both sides of the equation by V₂
p₂ = p₁ × V₁/V₂
p₁ = 485 kPa; V₁ = 648 mL
p₂ = ?; V₂ = 0.15 L = 150 mL Calculate p₂
p₂ = 485 × 648/150
p₂ = 2100 kPa
