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Draw the predominant product(s) of the following reactions including stereochemistry when it is appropriate. CH3CH2 C C CH3 H2O/H2SO4/HgSO4

Answers

Answer 1

Answer:

Answer:

Draw the predominant product(s) of the following reactions including stereochemistry when it is appropriate.

CH3CH2 C C CH3 H2O/H2SO4/HgSO4

Explanation:

The given compound is: pent-2-yne.

When it reacts with water, in presence of sulphuric acid and mercuric sulphate then a ketone is formed as shown below:

This reaction is an example of nucleophilic attack of water on carbon carbon triple bond.

The general mechanism of the reaction is hsown below:

Pent-2-yne reacts with water and form 3-pentanone.

The reaction is shown below:

Answer 2

Answer:

Final answer:

The reaction is the hydration of an alkene in an acidic environment, resulting in the formation of 2-butanol. This result is in accordance with Markovnikov's rule, which determines the position of the hydroxyl group in the resultant product.

Explanation:

The question refers to the acidity-catalyzed hydration of an alkene. In this case, you have an alkene CH3CH2 - CC - CH3 reacting in an acidic environment with water (H2O). The reactants have been exposed to H2O/H2SO4/HgSO4. In this reaction scenario, the acidic medium (H2SO4) and the water enact the role of a nucleophile and attack the alkene, thereby hydrating it.

The product of this reaction will be 2-butanol. Its formation is guided by Markovnikov's rule, which states that in the addition of a protic acid HX to an alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents. This rule is why the hydroxyl group (-OH) attaches itself to the 2nd carbon atom in the major (predominant) product.

Learn more about Acidity-catalyzed hydration here:

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Draw the structure that corresponds with the name: 3,5-dimethoxybenzaldehyde

Answers

Answer:

Structure is attached.

Explanation:

It is quite simple. The parent compound is benzaldehyde which is a benzene containing CHO functional group.

Now, starting numbering from CHO put methoxy (-OCH₃) groups at position 3 and 5.

After clamping a barrette to a ring stand you noticed that the set up is tippy and unstable what should you do to stabilize the set up?

Answers

Move the buret clamp to a ring stand with a larger base.

The entropy of a substance above absolute zero will always be: a. Negative
b. Positive
c. Neither Negative nor positive

Answers

i will be positive. just because it’s positive

If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, how many moles of Pb2+ were originally in the solution?

Answers

Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

Help me assiment will close after an hour

Answers

Answer :

(a) The molecular equation will be,

$CaCO_3(aq)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)$

(b) The complete ionic equation in separated aqueous solution will be,

$Ca^(2+)(aq)+CO_3^(2-)(aq)+2H^(+)(aq)+2Cl^(-)(aq)\rightarrow Ca^(2+)(aq)+2Cl^-(aq)+H_2O(l)+CO_2(g)$

(c) The net ionic equation will be,

$CO_3^(2-)(aq)+2H^(+)(aq)\rightarrow H_2O(l)+CO_2(g)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The molecular equation will be,

$CaCO_3(aq)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)$

(b) The complete ionic equation in separated aqueous solution will be,

$Ca^(2+)(aq)+CO_3^(2-)(aq)+2H^(+)(aq)+2Cl^(-)(aq)\rightarrow Ca^(2+)(aq)+2Cl^-(aq)+H_2O(l)+CO_2(g)$

In this equation, $Ca^(2+)\text{ and }Cl^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

(c) The net ionic equation will be,

$CO_3^(2-)(aq)+2H^(+)(aq)\rightarrow H_2O(l)+CO_2(g)$

Given that e = 9.0 v , r = 98 ω and c = 23 μf , how much charge is on the capacitor at time t = 4.0 ms

Answers

Let charge across the capacitor be Q, current through the circuit be I.
Voltage difference across the resistor = rI
Voltage difference across the capacitor = Q/c
Loop rule: net voltage change through a loop must be zero, so
9 = rI + Q/c. Since I = dQ/dt,
r dQ/dt + Q/c = 9
Solving, Q = 9c (1 - e^(t/rc)). Plug in the numbers from the problem for the numerical answer.

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