Answer:
<h2>Speed = 5.107 mile/hr</h2>
Explanation:
<h3>Given:</h3>
<u>distance</u> =50m = 0.0311 mile
<u>Time</u>= 21.93 seconds/3600 =0.00609 hour
<u>Speed</u><u>=</u>?
<h3>
Speed= Distance / Time </h3><h3>
= 0.0311 / 0.00609</h3><h3>
= 5.107 mile/hr</h3>
Answer:
okayyyt i wil taggg u ive seen some
In question 13 answer B is the correct one since a catalyst is a chemical that can be added to a reaction to increase the reaction rate without being consumed in the process. In this way, a catalyst accelerates a reaction by decreasing the energy barrier necessary for the reaction to occur or by changing the reaction mechanism. The most common types of catalysts are enzymes, acid base catalysts and heterogeneous catalysts, mostly composed of solids in which the reactants adhere.
In the reaction you have in the question, they use an acid catalyst, H3O+. This catalyst changes the reaction mechanism of 2-butene in 1-butene without being consumed, since it appears at the beginning and at the end of the transformation of 2 butene into 1-butene. The reaction mechanism would be the following
In question 14 the correct answer is option B since the law of velocity of a reaction will be governed by the slow step of the mechanism through which it occurs. In the slow step the chemical species have a harder time transforming because they need more energy to do it, then they will do it in a slower way and the reaction will take longer to occur. Therefore, the slow step is the one that will determine the total speed of the reaction and the speed law must be in terms of that process. The speed of the other two steps is so fast that it is negligible.
Answer:
Use a 10 mL pipet and a 50 mL volumetric flask, <em>twice</em>
Explanation:
Given those conditions, in order to have a solution that is 25 times less concentrated that the one previously prepared, we would need two dillution steps, as suggested by the problem.
If we take an aliquot of 10 mL and dilute it to a final volume of 50 mL, we prepare a solution that is 5 times less concentrated. And if we then take again an aliquot of 10 mL of that dilluted solution and dilute it to a final volume of 50 mL, we would have a solution 25 (5 times 5) times less concentrated than the original one.
In order to verify, we can calculate the resulting concentration and compare it with the original one:
Thus we can see that the obtained concentration is in fact 25 times lower than the original concentration.
