Rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122M.
A rate law indicates the rate of a chemical response depends on reactant concentration. For a response inclusive of the price regulation commonly has the form rate = ok[A]ⁿ, in which okay is a proportionality constant known as the fee regular and n is the order.
The charge of a chemical response is, perhaps, its maximum crucial asset because it dictates whether or not a reaction can arise all throughout an entire life. knowing the charge regulation, an expression concerning the price to the concentrations of reactants can assist a chemist to modify the response conditions to get an extra suitable rate.
half-life is the time taken for the radioactivity of a substance to fall to 1/2 its authentic cost whereas implies existence is the common life of all the nuclei of a particular risky atomic species.
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To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:
Then you add all of them together:
Therefore, the molar weight of glucose is 180.15 grams.
Answer:
Explanation:
If l = 3, the electrons are in an f subshell.
The number of orbitals with a quantum number l is 2l + 1, so there
are 2×3 + 1 = 7 f orbitals.
Each orbital can hold two electrons, so the f subshell can hold 14 electrons.
Explanation:
The molarity of a solution is defined like the number of moles of solute per liters of solution.
molarity = moles of solute/(volume of solution in L)
We know the volume of solution in L.
volume of solution = 0.65 L
To go from the mass of our solute in grams to moles we have to use its molar mass.
mass of NaCl = 63 g
molar mass of NaCl = 58.44 g/mol
moles of NaCl = 63 g * 1 mol/(58.44 g)
moles of NaCl = 1.078 moles
Finally we can find the molarity of the solution
molarity = moles of NaCl/(volume of solution)
molarity = 1.078 moles/(0.65 L)
molarity = 1.66 M
Answer: the molarity of the solution is 1.66 M.
We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams
