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8090 [49]
3 years ago
8

If a sample of DNA contained 20% cytosine, what percentage of guaninewould be in this sample and what percentage of adenine woul

d be in the sample
Chemistry
1 answer:
timofeeve [1]3 years ago
7 0
Guanine and Cytosine are complementary pairs in the DNA molecule (which means that Cytosine is bonded via hydrogen bonds to Guanine).  This is the same for Adenine and Thymine.  As such if a sample of DNA contains 20% Cytosine then it also contains approximately 20% Guanine.

The percentage of Thymine + Adenine would be 100% - 40% = 60%... Since this 60% should be equally shared between Thymine and Adenine then it implies that the percentage Adenine would be approximately 30%.
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Janice is given a mixture of alcohol and water. The teacher tells her that she can use temperature to separate these compounds.
Korolek [52]
Since Janice was given a mixture of alcohol and water, her teacher suggested that she use temperature to separate the two substances. The property demonstrated by the experiment is D. boiling. The boiling point refers to the temperature at which the liquid phase of the substance will turn into vapor. Water is known to boil at 100°C at atmospheric pressure while alcohols are generally known to have a boiling point lower than that of water. In this experiment, knowing that the two substances had a significant difference in boiling temperature was crucial to be able to separate them into their pure substances.
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3 years ago
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The equilibrium constant, Kc, for the following
umka21 [38]

Answer:

2CH2Cl2(g) Doublearrow CH4(g) + CCl4(g)

0.205 moles of CH2Cl2 is introduced. Let by the time an equilibrium is reached x moles each of CH4 and CCl4 are formed => remaining moles of CH2Cl2 are 0.205-x

i.e at equilibrium the concentration on CH2Cl2 is (0.205-2x) mol/L, CH4 is x mol/L, CCl4 is x mol/L

Now the equilibrium constant equation : K = [CH4][CCl4]/[CH2Cl2]^2 ([.] - stands for concentration of the term inside the bracket)

10.5 = x*x/(0.205-2x)^2

=> 10.5(4x^2-0.82x+0.042) = x^2

=>42x^2-8.61x+0.441=x^2

=>41x^2-8.61x+0.441 = 0

This is a Quadratic in x, solving for the roots, we get x = 0.0886 , x = 0.121

The second solution for x will lead 0.205-2x to become negative, so is an infeasible solution.

Therefore equilibrium concentrations of the products and reactants correspond to x=0.0886 and they are , [CH2Cl2] = 0.205-2*0.0886 =0.0278 mol/L , [CH4] = 0.0886 mol/L , [CCl4] = 0.0886 mol/L

4 0
2 years ago
According to coulomb’s law, what happens to the potential energy of two oppositely charged particles as they get closer together
ValentinkaMS [17]

Answer:

According to coulomb's law their potential energy will increase and the particles will repel each other.

8 0
2 years ago
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Are soft drink, solder and air homogeneous or heterogeneous?
NeTakaya
They are all Homogeneous
4 0
2 years ago
A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside leve
mixas84 [53]

Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

                  PV = nRT

or,           n = \frac{PV}{RT}

                 = \frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}

                  = 0.0052 mol

Also,  No. of moles = \frac{mass}{\text{molar mass}}

               0.0052 mol = \frac{mass}{2 g/mol}

                  mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at 25^{o}C

                P_{\text{water vapor}} = 24 mm Hg

                                = \frac{24}{760} atm

                                = 0.03158 atm

Now,   P = \frac{756}{760} - 0.03158

              = 0.963 atm

Hence,   n = \frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}

                 = 0.0056 mol

So, mass of H_{2} = 0.0056 mol × 2

                         = 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

      Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100

                              = \frac{0.01013 g}{0.0104 g} \times 100            

                              = 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

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