Answer: 0.0069L
Explanation:
2H2O(l) ---->O2(g) + 4H+(aq) + 4e-
no of moles= it/eF
NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)
Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)
= 0.0002798 moles= 2.798x 10 ^-4moles
Using ideal gas equation,
P V = n R T
Where, P is the pressure,
V is the volume,
n is the number of moles,
R is the gas constant, and T is the temperature
We have, 1 bar = 0.986923 atm
Substituting the values,
V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L
Volume of O2 produced = 0.0069L
The molecule that contains the fewest number of Hydrogen atoms would be B. Al(OH)3. It only has 3 Hydrogen atoms.
The answer is B
Homozygous mean the same and you know that the two alleles will be the same (either BB or bb) and receive is usually the lower case set of alleles
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
Answer:
The required mass to prepare 2.5 L of 1.0 M NaOH solution is 100 g
Explanation:
We do this by preparing the equation:
Mass = concentration (mol/L) x volume (L) x Molar mass
Mass = 1.0 M x 2.5 L x 40 g/mol
Mass = 100 g
