Answer:
Tom is correct. The rate of reaction of Crystal violet and NaOH is first order with respect to NaOH, hence, a higher concentration of NaOH corresponds to a higher rate of reaction; a faster reaction.
This means the speed of cleaning depends on the concentration of the lye used.
Explanation:
The reaction between Crystal violet and NaOH, represented as
CV⁺ + OH⁻ → CVOH
It is a reaction that is know to turn the violet colour of the crystal violet colourless.
The rate of the reaction is also known to be second order; first order with respect to Crystal violet and first order with respect to NaOH.
This means that the rate of reaction is directly proportional to the concentration of NaOH provided all other parameters such as the rate constant and the concentration of Crystal violet are constant.
Hence, the reaction becomes faster with an increased concentration of NaOH.
So, Tom is right, concentrated lye solution would remove the stain faster.
Bob is wrong.
Hope this Helps!!!
Answer:
b,d
Explanation:
An ion is an atom with a difference in the amount of protons and elektrons.
The people of Finland, who are secluded to some degree from the rest of the world by water, develop certain diseases due to the lack of genetic material from other ethnicities and races.
Physical barriers prevent fish from one stream from mating with fish from another stream, leading to a less varied gene pool among those fish. As time passes, the fish become unable to successfully mate with other groups.
A mountain range prevents two types of goat from mating, causing the gene pool to become less varied.<span>
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Answer:
C
Explanation:
Wind is geological therefore it is geological weathering
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
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2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
