The reaction is not balanced
<h3>Further explanation</h3>
Given
Reaction
2Fe(s)+3O₂(g)⇒2Fe₂O₃(s)
Required
The number of atoms
Solution
In a balanced chemical equation, the number of atoms in the compound that reacts (the reactants and products) will have the same number
Reactants : Fe(s)+O₂(g)
Fe = 2 atoms
O = 3 x 2 = 6 atoms
Products : Fe₂O₃(s)
Fe = 2 x 2 = 4 atoms
O = 2 x 3 = 6 atoms
The reaction is not balanced because the number of Fe atoms is not the same
The balanced reaction should be:
4Fe(s)+3O₂(g)⇒2Fe₂O₃(s)
Answer:
0.6749 M is the concentration of B after 50 minutes.
Explanation:
A → B
Half life of the reaction = 
Rate constant of the reaction = k
For first order reaction, half life and half life are related by:
Initial concentration of A = ![[A]_o=0.900 M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.900%20M)
Final concentration of A after 50 minutes = ![[A]=?](https://tex.z-dn.net/?f=%5BA%5D%3D%3F)
t = 50 minute
![[A]=[A]_o\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_o%5Ctimes%20e%5E%7B-kt%7D)
![[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}](https://tex.z-dn.net/?f=%5BA%5D%3D0.900%20M%5Ctimes%20e%5E%7B-0.02772%20min%5E%7B-1%7D%5Ctimes%2050%20minutes%7D)
[A] = 0.2251 M
The concentration of A after 50 minutes = 0.2251 M
The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M
0.6749 M is the concentration of B after 50 minutes.
The balanced chemical reaction is given as follows:
<span>2 KClO3(s) → 2 KCl(s) + 3 O2(g)
The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:
</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
Answer:
No
Explanation:
Depending on the mass of the horse and the speed, velocity will change.
Answer:
N2 + 3H2 ———> 2NH3
As we know 1000 grams ammonia is 58.82 moles so according to unitary method,
2 mole NH3 formed by 1 mole N2 hence 58.82 NH3 will be given by 29.41 moles N2.
No. Of moles = given mass/molar mass
Implies that
Mass of nitrogen required = 29.41*28 = 823.48 grams.
Explanation:
