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3 years ago

Draw the carboxylic acid produced from the acid hydrolysis of butyl acetate.

Chemistry

1 answer:

Nat2105 [25]3 years ago

7 0

Answer:

Acetic or ethanoic acid.

Explanation:

Hello,

Butyl acetate's hydrolisis is shown below:

$CH_3COOCH_2CH_2CH_2CH_3+H_2O-->CH_3COOH+CH_3CH_2CH_2CH_2OH$

Whereas $CH_3COOCH_2CH_2CH_2CH_3$ is the butyl acetate, $CH_3CH_2CH_2CH_2OH$ butanol and $CH_3COOH$ the required produced acid, acetic or ethanoic acid (attached picture).

Best regards.

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Ii) The valency of which the element is zero.<br>a) lead. b) bromine c) helium. d) nitrogen

stira [4]

Answer:

Helium, Neon, Radon and Argon have zero valency because they have their outer shell completely filled and are chemically inert gas. Was this answer helpful

Explanation:

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6 0

1 year ago

Read 2 more answers

How to balance lead and silver acetate yields lead (||) acetate and silver

arsen [322]

The given elements put into an equation using their symbols are as follows:
Pb + $AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb + $AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb + $2AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb + $2AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + 2Ag

That is your final equation

The coefficients are 2 + 2 = 1 + 2

3 0

3 years ago

Titration Volume & Concentration

polet [3.4K]

Answer:

18,1 mL of a 0,304M HCl solution.

Explanation:

The neutralization reaction of Ba(OH)₂ with HCl is:

2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O

The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:

$0,0171L*\frac{0,161moles}{L}$ = 2,7531x10⁻³moles of Ba(OH)₂

By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:

$2,7531x10^{-3}molBa(OH)_{2}*\frac{2molHCl}{1molBa(OH)_{2}}$ = 5,5062x10⁻³moles of HCl.

The volume of a 0,304M HCl solution for a complete neutralization is:

$5,5062x10^{-3}molHCl*\frac{1L}{0,304mol}$ = 0,0181L≡18,1mL

I hope it helps!

4 0

3 years ago

How can I balance this equation? TELL ME HOW you did it and all the steps and you will be the brainliest.

lianna [129]

Answer:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Explanation:

To balance, start of with the groups that are common on both sides of the reaction equation;

In this case, these are the SO4 groups treating them as single units;

There are 3 on the right side and 1 on the left side, so we put 3 in front of the H2SO4 to balance this first;

Next, deal with the Cr, there are 2 Cr on the right side and 1 on the other side, so we put a 2 in front of the H2CrO4 to balance that;

Thirdly, we notice the C are already balanced as there are 2 on each side so this is fine;

Lastly, we can deal with the O and H;

Bearing in mind the numbers that are in front of the molecules now from prior balancing, there are 9 O and 16 H on the left side and 3 O and 5 H on the right;

7 H2O on the right side would balance the O, but gives us 18 H, which is 2 too many H;

If we were to put 2 in front of the two organic molecules (the ones with C) on either side, we would balance the O by having 6 H2O, but this gives 2 fewer H than necessary;

In order for the H to balance, we need to have 13/2 (or 6.5) H2O, which means we need 3/2 (or 1.5) in front of each organic molecule;

Since, it is not sensible to have 13/2 water molecules or 3/2 organic molecules, we can just multiply everything by 2;

Thus we end up with:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Rules of thumb:

- When there are common chemical groups (e.g. SO4) on both sides of a reaction equation, treat them as single units

- Start of with balancing these common groups

- Thereafter, balance the atoms that appear in only one reactant and one product

- Proceed to balance the atoms that appear in more than one reactant or product

- Typically, you should deal with the O and H last

4 0

3 years ago

How many moles are in 17.2g k2s

Natalija [7]

Answer: 0.156 mol

Explanation:

To find the moles of 17.2 g K₂S, we need to know the molar mass to convert.

$17.2g*\frac{mol}{110.256 g} =0.156 mol$

3 0

4 years ago