Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V
Yes... that is correct.
CH4 is methane so the coefficent in front of it would double the number of atoms of each element
Answer:
D) There must be equal number of atoms of each elements on both sides of equation.
Explanation:
The balancing equation must have equal number of atoms of each elements on both sides of equation.
The balance equation shoes mass is conserved thus followed the law of conservation of mas.
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
Explanation:
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
For example:
In given photosynthesis reaction:
6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂
there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.
The answer is 5.32 × 10²³ molecules
<span>Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance:
</span>6.023 <span>× 10²³ units per 1 mole
We have 0.883 moles.
If 1 mole has </span>6.023 × 10²³ molecules, 0.883 moles will have x molecules:
1 mole : 6.023 × 10²³ molecules = 0.883 moles : x
x = 6.023 × 10²³ molecules * 0.883 moles : 1 mole = 5.32 × 10²³ molecules
<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
Or,
where,
= osmotic pressure of the solution = 15.5 mmHg
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)
Volume of solution = 6.5 mL
R = Gas constant = 
T = temperature of the solution = ![25^oC=[273+25]=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5D%3D298K)
Putting values in above equation, we get:
Hence, the molar mass of the insulin is 6087.2 g/mol
