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Usimov [2.4K]
3 years ago
14

Provide an appropriate alkyne starting material A and intermediate product B. Omit byproducts. The number of carbon atoms in the

starting material should be the same as in the final product.
Chemistry
1 answer:
nadezda [96]3 years ago
6 0

Answer:

C4H6 + 2HBr = C4H8Br2

Explanation:

Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of CnH2n−2. They are unsaturated hydrocarbons.

Furthermore, Halogenation reaction of any alkyne such as but-2-yne with any of the hydro halogenated compounds such as HBr, HCl, HF or HI will give an alkane that will have the same number of carbon atoms in the starting material and in the final product.

Halogenation reaction is a reaction that occurs when one or more halogens are added to a substance. Halogens is the seventh column in the periodic table and it include fluorine, chlorine, bromine, iodine, and astatine.

You might be interested in
A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The compone
AysviL [449]

Answer:

0,07448M of phosphate buffer

Explanation:

sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:

Na₂HP + HCl ⇄ NaH₂P + NaCl <em>(1)</em>

NaH₂P + HCl ⇄ H₃P + NaCl <em>(2)</em>

The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:

0,01862L of HCl×\frac{0,1000mol}{L}= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.

The concentration is:

\frac{1,862x10^{-3}moles}{0,02500L} = <em>0,07448M of phosphate buffer</em>

<em></em>

I hope it helps!

7 0
3 years ago
2. The reaction of zinc with nitric acid was carried out in a calorimeter. This reaction caused the temperature
snow_lady [41]

We want to solve Q = mcΔT for the liquid water; its change in temperature will tell us the amount of thermal energy that flowed out of the reaction. The specific heat, c, of water is 4.184 J/g °C.

Q = (72.0 g)(4.184 J/g °C)(100 °C - 25 °C) = 22593.6 J

Q ≈ 2.26 × 10⁴ J or 22.6 kJ (three significant figures).

3 0
3 years ago
Metals bond with halogens to form colorless metal halides. During an experiment, bromine water was added to a solution of potass
Anvisha [2.4K]

Answer:

Bromine displaced iodide to form iodine

Explanation:

In this experiment, liquid bromine was allowed to react with a solution of potassium iodide.

Initially, we had a colorless KI solution which was mixed with light orange Br_2. If we add a light orange solution to a colorless solution, liquid bromine will get diluted and we'll see a slight decrease in its color intensity, however, if no reaction takes place, no color change will occur.

In the experiment, however, we examine a formation of a deep brown solution. Remember that iodine solution would form a brown solution.

As a result, bromine displaced iodide anion from potassium iodide:

2~KI~(aq) + Br_2~ (l)\rightarrow 2~KBr~(aq) + I_2~(aq)

8 0
3 years ago
How many molecules are in 79g of fe2o3
horrorfan [7]
Convert grams —> mols and then mols —> atoms

We know that there are 6.02 x 10^23 atoms/mol

And we know that there are about 160 grams of fe2o3 per mol

So (79g fe2o3)/(160 g/mol) = .49 mol fe2o3

Now we use avogadro’s number to do

(.49 mol fe2o3)/(6.02 x 10^23 atoms/mol) = the answer.

I’ll leave the easy math to you.
7 0
3 years ago
The pH of a 0.29 M solution of carbonic acid (H2CO3) is measured to be 3.44. Calculate the acid dissociation constant Ka of carb
arsen [322]

Answer: 4.55x10^-7

Explanation:

H2CO3 + H20 <==> H3O+ + HCO3-

Desigining an ICE table, we have:

Initial conc. of H2CO3 = 0.29 M

Initial conc. of H3O+ = 0

Initial conc. of HCO3- = 0

Change in conc. of H2CO3 = - x

Change in conc. of H3O+ = x

Change in conc. of HCO3- = x

Equilibrium conc. of H2CO3 = 0.29 - x

Equilibrium conc. of H3O+ = x

Equilibrium conc. of HCO3- = x

but pH = 3.44

pH = - log[H30+]

[H30+] = 10^-3.44 = 3.63x10^-4

[H30+] = 3.63x10^-4

Ka = [H3O+].[HCO3-] / [H2CO3]

[H30+] = x = 3.63x10^-4

[HCO3-] = 3.63x10^-4

[H2CO3] = 0.29 - x = 0.29 - 3.63x10^-4 = 0.289637

Ka = (3.63x10^-4)(3.63x10^-4)/0.289637 = 4.55x10^-7

6 0
3 years ago
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