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3 years ago

Mr. Jenson's class is observing the growth of one kind of pea plant.

Chemistry

1 answer:

N76 [4]3 years ago

4 0

Answer:

The answer would be C. Giving the pea plants a different amount of water in each one can help you observe how much water produces the most peas in a pea pod.

The answer is C to give all of the pea plants different amounts of water and count the number of peas that the plants produce

Explanation:

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In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is

V125BC [204]

Answer:

the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

Explanation:

Given that:

mass of diphenylacetylene $(C_{14}H_{10})$ = 0.5297 g

Molar Mass of diphenylacetylene $(C_{14}H_{10})$ = 178.21 g/mol

Then number of moles of diphenylacetylene $(C_{14}H_{10})$ = $\frac{mass}{molar \ mass}$

= $\frac{0.5297 \ g }{178.24 \ g/mol}$

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = Heat absorbed by $H_2O$ + Heat absorbed by the calorimeter

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = msΔT + cΔT

= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = 20598.87 J

Heat liberated by 1 mole of diphenylacetylene $(C_{14}H_{10})$ will be = $\frac{20598.87 \ J}{0.002972 \ mol}$

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

3 0

3 years ago

Read 2 more answers

An archaeologist discovered some charcoal at an excavation site. He sent the samples to a lab for C-14 dating, knowing that the

klio [65]

Here we assume that disintegration has occur so the normal t½ formular will not be used
mathematically we..⅛×5600 =700
therefore it is about 700years option A

7 0

3 years ago

Encountering problems during separation/purification is a very common issue in organic synthesis, but these steps are crucial to

Rus_ich [418]

True is the correct answer! your welcome! :)

8 0

3 years ago

Find the volume of 20g of H₂ at STP<br><br>

ANEK [815]

Answer:224

Explanation:

We should answer it with Stoichiometry

We say: 20 g H2× (1 mol/ 2g)× ( 22.4 lit/ 1 mol) = 224

Means: we have 20 grams and every 2g H2, equals to 1 mol of it and every 1 mol of H2, equals to 22.4 lit( because of STP)

hope you got this:)

6 0

3 years ago

Compound a is an alkene that was treated with ozone (followed by dms) to yield only 4-heptanone. Draw the major product that is

olga nikolaevna [1]

Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.

Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.

The reaction is as shown in the image.

The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid $H_3O^{+}$ is shown in the image.

m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.

5 0

3 years ago