Answer: The reaction order with respect to A is m
Explanation:
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.
For the given reaction:
![Rate=k[A]^m[B]^n](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5Em%5BB%5D%5En)
In this equation, the order with respect to each reactant is not equal to its stoichiometric coefficient which is represented in the balanced chemical reaction.
Hence, this is not considered as an elementary reaction.
Order with respect to A = m
Order with respect to B = n
Overall order = m+n
Thus order with respect to A is m.
Answer:
5.231 L.
Explanation:
- Molarity is the no. of moles of solute per 1.0 L of the solution.
<em>M = (no. of moles of KCl)/(Volume of the solution (L))</em>
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M = 6.5 M.
no. of moles of solute = 34.0 mol,
Volume of the solution = ??? L.
∴ (6.5 M) = (34.0 mol)/(Volume of the solution (L))
∴ (Volume of the solution (L) = (34.0 mol)/(6.5 M) = 5.231 L.
Answer:
1. All red calves i.e. RR
2. All roan calves i.e RW
3. 2 red calves (RR) and two roan calves (RW)
Explanation:
According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):
1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR
2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW
3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).
Plants that have nigrogen fixing bacteria in their roots are called
legumes.
Answer:
90.3 L
Explanation:
Given data:
Volume of water produced = 77.4 L
Volume of oxygen required = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
It is known that,
1 mole = 22.414 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
There are 6 moles of water = 6×22.414 = 134.5 L
Now we will compare:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.
