Answer:
see explanation below
Explanation:
You are missing the reaction scheme, but in picture 1, I found a question very similar to this, and after look into some other pages, I found the same scheme reaction, so I'm gonna work on this one, to show you how to solve it. Hopefully it will be the one you are asking.
According to the reaction scheme, in the first step we have NaNH2/NH3(l). This reactant is used to substract the most acidic hydrogen in the alkine there. In this case, it will substract the hydrogen from the carbon in the triple bond leaving something like this:
R: cyclopentane
R - C ≡ C (-)
Now, in the second step, this new product will experiment a SN2 reaction, and will attack to the CH3 - I forming another alkine as follow:
R - C ≡ C - CH3
Finally in the last step, Na in NH3 are reactants to promvove the hydrogenation of alkines. In this case, it will undergo hydrogenation in the triple bond and will form an alkene:
R - CH = CH - CH3
In picture 2, you have the reaction and mechanism.
<u>36 ml of NaOh and</u><u> 464 ml</u><u> of </u><u>HCOOH</u><u> would be enough to form 500 ml of a buffer with the same pH as the buffer made with </u><u>benzoic acid </u><u>and NaOH.</u>
What is benzoic acid found in?
- Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
- Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.
Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.
Amount of moles of NaOH -2 × 0.025 = 0.05 mol
Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol
In this case, we can calculate the pH produced by the buffer of these two reagents, as follows
We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows
Now we must solve the equation above. This will be done using the following values
With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.
NaOH volume
( 0.5 - 0.464)L
0.036L .................... 36ml
HCOOH volume
500 - 36 = 464mL
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Answer:
B. The air pressure would increase 2X
Explanation:
Pressure = density × acceleration due to gravity × height
p = ρ × g × h
Pressure is force per unit area
p = F/a
Density is mass per unit volume
Pressure is directly proportional to density hence, any change in the density of a body directly affects the change in pressure of the body and vice versa.
The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.
First-order reactions are very commonplace. we have already encountered examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to present t-butanol. every other reaction that famous obvious first-order kinetics is the hydrolysis of the anticancer drug cisplatin.
The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.
Rate constant K₁ = 0. 543 /s
T₁ = 25°C
Activation energy Eₐ = 75. 9 k j/mol.
T₂ = 32. 3 °C.
K₂ =?
formula;
log K₂/K₁= Eₐ /2.303 R [1/T₁ - 1/T₂]
putting the value in the equation
K₂ = 0.343 /s
Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s
The specific rate steady is the proportionality consistent touching on the fee of the reaction to the concentrations of reactants. The fee law and the specific charge consistent for any chemical reaction should be determined experimentally. The cost of the charge steady is temperature established.
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Answer:
I think cold front if not than its c
