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2 years ago

What are the structural features of purines and pyrimidines?

1 answer:

7 0

The purines in DNA are adenine and guanine, the same as in RNA. The pyrimidines in DNA are cytosine and thymine; in RNA, they are cytosine and uracil. Purines are larger than pyrimidines because they have a two-ring structure while pyrimidines only have a single ring

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How many moles are in 3.0g of A1203?

vfiekz [6]

Answer:

0.065 moles

Explanation:

4 0

3 years ago

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

Answer: The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago

Which accurately describes nonmetals

Bumek [7]

Answer:

Most nonmetals are solids, but some are gaseous or liquid. All nonmetals are solid unless they bond with a metal.

Explanation:

ANSWER: LOOK IT UP IN YO DICtionary

- williams got dem guns

8 0

2 years ago

What is the molality of a solution of water and KCl if the freezing point of the solution is –3°C? (Kf = 1.86°C/m; molar mass of

Elina [12.6K]

Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. Calculations are as follows:

ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg

4 0

3 years ago

Read 2 more answers

Mercury has an atomic mass of 200.59 amu. calculate the mass of 3.0 x 10^10 atoms

Licemer1 [7]

To determine mass of the given number of atoms of mercury, we need a factor that would relate the number of atoms to number of moles. In this case, we use the Avogadro's number. It is a number that represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. The number of units could be atoms, molecules, ions or electrons. To convert into mass, we use the given amu of mercury since it is equal to grams per mole. We calculate as follows:

3.0 x 10^10 atoms ( 1 mol / 6.022 x 10^23 atoms ) ( 200.59 g / 1 mol ) = 9.99x10^-12 g Hg

5 0

2 years ago