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3 years ago

#PLATOFAM

2 answers:

4 0

Ketones and aldehydes are simple compounds that contain a carbonyl group (a carbon-oxygen double bond). They are considered "simple" because they do not have reactive groups like −OH or −Cl attached directly to the carbon atom in the carbonyl group, as in carboxylic acids containing −COOH.

So your answer would be D.).

I hope that this helps. :)

My name is Ann [436]3 years ago

3 0

The organic compound ketone contains which functional group?

D.) carbonyl

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You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula: $\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n$

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345$

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0

3 years ago

In a reaction, 25 grams of reactant

Vsevolod [243]

Answer:

D. 15g

Explanation:

The law of conservation of mass states that, in a chemical reaction, mass can neither be created nor destroyed. This means that the amount of matter in the elements of the reactants must be equal to the amount in the resulting products.

In this question, 25 grams of a reactant AB, was broken down in a reaction to produce 10 grams of products A and X grams of product B. According to the law of conservation of mass, the mass of the reactant must be equal to the total mass of the products. This means that 25 grams must also be the total mass of both products in this reaction. Hence, if product A is 10 grams, product B will be 25 grams - 10 grams = 15 grams.

Therefore, product B must be 15 grams in order to form a total of 25 grams when added to the mass of product A. This will equate the mass of the reactant AB and fulfill the law of conservation of mass.

8 0

3 years ago

Calculate the molality of an ethylene glycol (C2H6O2) soltuion, that has a molarity of 2.07 M. The density of the solution is 1.

vlada-n [284]

Answer:

2.03

Explanation:

Let's assume we have 1 L of the solution:

There would be 2.07 ethylene glycol moles.

The solution would weigh (1000 mL * 1.02 g/mL) = 1020 g.

With that information we can calculate the molality:

molality = moles of solute / kg of solvent

molality = 2.07 moles / (1020 ÷ 1000) = 2.03 m

Keep in mind that this is only an estimate, as we used the kg of the solution and not of the solvent.

3 0

3 years ago

PLEASE HELP!!!

trasher [3.6K]

Answer:

(a) oxygen

(b) 154g (to 3sf)

Explanation:

mass (g) = moles × Mr/Ar

note: eqn means chemical equation

(a)

moles of P = 84.1 ÷ 30.973 = 2.7152 moles

moles of O2 = 85÷2(16) = 2.65625 moles

Assuming all the moles of P is used up,

moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)

moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)

therefore there is insufficient moles of O2 and the limiting reactant is oxygen.

(b)

moles of P2O5 produced

= 2/5 (according to eqn) × 2.7152

= 1.08608moles

mass of P2O5 produced

= 1.08608 × [ 2(30.973) + 5(16) ]

= 154.164g

= approx. 154g to 3 sig. fig.

(c)

% yield = actual/theoretical yield × 100%

= 123/154 × 100%

= 79.870%

= approx. 79.9% (to 3sf)

4 0

3 years ago

Solid carbon is known to react with oxygen gas to produce carbon dioxide. If a mass of 1.2g of carbon is burned in oxygen, 4.4g

Elden [556K]

Now ,

C + O2 → CO2

According to above equation, 1 mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.Thus this implies that 12 g of carbon reacts with 32 g of O2 to produce 44 g of CO2.

No of moles = mass of the substance/molecular mass of the substance.

In this case 1.2 g of carbon reacts with "x "g of O2 to produce 4.4 g of CO2.

No of moles of carbon in this case = 1.2÷ 12 = 0.1 moles.

No of moles of carbon dioxide formed = 4.4÷44 =0.1 moles

Thus already discussed above, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. Hence to produce 0.1 mole of CO2 ,0.1 mole of carbon needs to react with 0.1 mole of oxygen.

Also number of moles of O2 = mass of O2÷ molar mass of O2

Substituting number of moles of O2 as 0.1 we get

mass of O2(x) = Number of moles of O2 × Molar mass of O2

Mass of O2 (x) = 0.1 × 32= 3.2 g

Thus mass of 3.2 g O2 reacts with 1.2 g of CO2 to produce 4.4 g of CO2.

7 0

3 years ago