Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
Answer:
have you tasted acid?
Explanation:
also the taste of salt is sour.
Answer:
18.94%.
Explanation:
- The decay of carbon-14 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
- The integration law of a first order reaction is:
<em>kt = ln [A₀]/[A]</em>
k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.
t is the time = 13,750 years.
[A₀] is the initial percentage of carbon-14 = 100.0 %.
[A] is the remaining percentage of carbon-14 = ??? %.
∵ kt = ln [Ao]/[A]
∴ (1.21 x 10⁻⁴ year⁻¹)(13,750 years) = ln (100.0%)/[A]
1.664 = ln (100.0%)/[A]
Taking exponential for both sides:
5.279 = (100.0%)/[A]
<em>∴ [A]</em> = (100.0%)/5.279 = <em>18.94%.</em>
Answer:
18 inorganic phosphates and 18 pyruvates.
Explanation:
Glycolysis is the pathway involved in the metabolism of sugar. It is enzyme catalyze and converts glucose into pyruvate and a Hydrogen ion.
The free energy released are the used to form high-energy molecules such as ATP and NADH.
The net ATP production of Glycolysis involving a molecule of Glucose are 2 ATPs ,2 NADH and 2 pyruvates.
If 9 molecules of glucose enter glycolysis,then it has to be multiplied by 9 to give 18 pyruvates and 18 net ATPs
Answer:
In the option(A) moles of HCl left are 0.100 moles which is wrong, making the option incorrect.
Explanation:
Moles of HCl = n
Molarity of HCl = 1.0M
Volume of HCl solution = 30.0 mL = 0.030 L (1 mL = 0.001L)
Moles of Fe = 
According to recation , 1mol of Fe reacts with 2 mol HCl. Then 0.01 mole of Fe will recat with :
of HCl
This means that HCl uis in excess , hence excessive reagent.
Moles of HCl left unreacted :
= 0.030 mol - 0.020 mol = 0.010 mol
But in the option moles of HCl left are 0.100 moles which is wrong, making the option incorrect.
