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3 years ago

The nitrogenous bases cytosine and thymine are

1 answer:

7 0

Purines are adenine and guanine. So the answer should be D (Pyrimidines). besides, phosphate group and pentose sugar are a part of a nucleotide.

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2.Please study the figures below which illustrate the steps in the chemical reaction of fluoromethane (CH3F) with a hydroxide io

Monica [59]

1) The names of the molecules

2) Write the chemical equation

$OH^-+CH_3F\rightarrow CH_3OH+F^-$

3) Balance the chemical equation

1 fluoride in the reactants and 1 fluoride in the products

1 carbon in the reactants and 1 carbon in the products

4 hydrogens in the reactants and four hydrogens in the products

1 oxygen in the reactants and 1 oxygen in the products

Since the number of atoms in the reactants is the same that the number of atoms in the products, the chemical equation is balanced.

7 0

1 year ago

The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm.

Vladimir [108]

Answer:

The molecular formula = $C_{6}H_{6}$

Explanation:

Given that:

Mass of compound, m = 0.145 g

Temperature = 200 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (200 + 273.15) K = 473.15 K

V = 97.2 mL = 0.0972 L

Pressure = 0.74 atm

Considering,

$n=\frac{m}{M}$

Using ideal gas equation as:

$PV=\frac{m}{M}RT$

where,

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the values in the above equation as:-

$0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15$

$M=78.31\ g/mol$

The empirical formula is = $CH$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 1 = 13 g/mol

Molar mass = 78.31 g/mol

So,

Molecular mass = n × Empirical mass

78.31 = n × 13

⇒ n ≅ 6

The molecular formula = $C_{6}H_{6}$

6 0

3 years ago

True or false all countries are legally bound to survey and monitor for disease within their bodies

ra1l [238]

Answer:

False..........

7 0

3 years ago

Explain the importance of carbon's ability to form covalent bonds in straight chains, branched chains, or rings.

viktelen [127]

Carbon is the element at the heart of all organic compounds, and it is such a versatile element because of its ability to form straight chains, branched chains, and rings. Because these chains and rings can have all sorts of different functional groups in all sorts of different ways (giving the compond all sorts of different physical and chemical properties), carbon's ability to form the backbone of these large structures is critial to the existence of most chemical compounds known to man. Above all, the organic molecules crucial to the biochemical systems that govern living organisms depend on carbon compounds.

8 0

3 years ago

At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz

Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P° (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles: n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

Hope this helps

7 0

3 years ago