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gulaghasi [49]
3 years ago
14

Which element is present in all organic compounds?

Chemistry
2 answers:
Mkey [24]3 years ago
5 0
The anser is 1, carbon
Karolina [17]3 years ago
5 0
Carbon is a basic element in all organic compounds, so (1) is the best choice.
Hope this would help:))
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True or false
Ivanshal [37]

Answer:

true because of the elements

4 0
3 years ago
The meselson-stahl experiment demonstrated that DNA replication is semi conservative in the figure semi conservative replication
Vlad1618 [11]

Answer:

The correct answer will be option-D.

Explanation:

The semi-conservative mode of replication is the common method of replication used by mostly organisms on earth which were demonstrated by the Meselson and Stahl in 1958.

They used the isotopes of the nitrogen which contains extra neuron which makes the nitrogen-heavy and cultured them in bacteria

The experiment concluded that newly replicated double stranded DNA contains one newly synthesized DNA strand which is complementary to another parental strand. Thus new DNA molecule has one parental strand while other newly synthesized strand.

This is best illustrated by the option-D in the figure and therefore is the correct answer.

7 0
3 years ago
Read 2 more answers
Determine the rate of a reaction that follows the rate law rate = k a m b i where
ikadub [295]
(missing part of your question):
when we have K = 1 x 10^-2 and [A] = 2 M & [B] = 3M & m= 2 & i = 1
So when the rate = K[A]^m [B]^i
and when we have m + i = 3 so the order of this reaction is 3 So the unit of K is L^2.mol^-2S^-1
So by substitution:
∴ the rate = (1x 10 ^-2 L^-2.mol^-2S^-1)*(2 mol.L^-1)^2*(3mol.L^-1)
                 = 0.12 mol.L^-1.S^-1
6 0
3 years ago
The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

3 0
3 years ago
2. Fe(s) + O2(g) → Fe3O4(s) a. When 13.54 g of O2 is mixed with 12.21 g of Fe, which is the limiting reactant? b. What mass in g
julsineya [31]

The percent yield shows the extent to which the reactants are converetd into products . The limiting reactant is used up in the reaction.

<h3>What is a limiting reactant?</h3>

A limiting reactant is the reactant that is in the least amount in the system.  Now;

Number of moles of Fe =  12.21 g/56 g/mol = 0.22 moles

Number of moles of O2 = 13.54 g/32 g/mol = 0.42 moles

Balanced reaction equation;

3Fe(s) + 2 O2(g) = Fe3O4(s)

If 3 moles of Fe reacts with 2 mole of O2

0.22 moles of Fe reacts with  0.22 moles * 2 mole/3 moles = 0.15 moles

Hence, Fe is the limiting reactant

If 3 mole of Fe produces 1 mole of  Fe3O4(s)

 0.22 moles  of O2   produces   0.22 moles * 1 mole/3 moles of  Fe3O4(s) = 0.1073 moles

Mass of  Fe3O4(s) =0.1073 moles  * 232 g/mol =16.9 g

Number of moles of excess reactant = 0.42 moles - 0.15 moles = 0.27 moles

Mass of excess reactant = 0.27 moles * 32 g/mol = 8.64 g

percent yield = 15.88 g/16.9 g * 100/1

= 93.4%

Learn more about percent yiled: brainly.com/question/13463225

8 0
2 years ago
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