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Rate of Reaction ca be increased by applying following changes,
1) Increase in Temperature:
Increase in Temperature results in the increase of Kinetic Energy, Hence, velocity of reactants increases and rate of collisions increases. Thus, more reactants colloid with enough energy to change into products in given time.
2) Decrease in Activation Energy:
Activation energy is the minimum amount of energy required by reactants to cross the energy barrier to convert into products. Hence, activation energy is decreased by using catalyst.
3) Concentration of Reactants:
The greater the concentration of reactants, greater is the number of reactants per volume. More reactants in less area increase the probability of collision. Hence, collisions increase and rate of reaction increases.
4) Increasing Surface Area of Reactant:
Greater the surface area greater is the rate of reaction.
Because I can't see your options, all I can say is to look for the one whose numbers can't be simplified any further
Answer:
1.395M NaOH
Explanation:
Sodium hydroxide, NaOH, reacts with nitric acid, HNO3, as follows:
NaOH + HNO3 → NaNO3 + H2O
<em>Where 1mol of NaOH reacts with 1mol of HNO3</em>
To solve this question we must find the concentration of the titrant. With the concentration and the needed acid we can find the moles of HNO3 added = moles NaOH in the solution. With the moles of NaOH and its volume we can find its concentration as follows:
<em>HNO3 concentration:</em>
10.00mol/L HNO3 * (125.0mL/500.0mL) = 2.500M HNO3
<em>Moles HNO3 = Moles NaOH:</em>
16.74mL = 0.01674L*(2.500mol/L) = 0.04185 moles HNO3 = Moles NaOH
<em>Concentration NaOH:</em>
0.04185 moles / 0.0300L =
1.395M NaOH
Answer:
8
Explanation:
From the question given above, the following data were obtained:
t–butyl ion = (CH₃)₃C⁺
Number of valence electron =?
The valence electron(s) talks about the combining power of an element or compound as the case may be.
Considering the t–butyl ion, (CH₃)₃C⁺ we can see that it has a charge of +1 indicating that it has given out 1 electron to attain the stable octet configuration which has a valence electrons of 8. Thus, the valence electron of t–butyl ion, (CH₃)₃C⁺ is 8
