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3 years ago

Nucleotides have a phosphate group attached at the ___ carbon atom of the sugar.

Chemistry

1 answer:

Tanzania [10]3 years ago

6 0

Answer:

Nucleotides have a phosphate group attached at the 5' carbon atom of the sugar.

Explanation:

Phosphate group is attached to the sugar molecule in the place of -OH.

You might be interested in

How many grams are in 4.5 x 10^22 molecules of Ba(NO2)2

NNADVOKAT [17]

Answer:

17 g Ba(NO₂)₂

General Formulas and Concepts:

Chemistry

Stoichiometry
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

4.5 × 10²² molecules Ba(NO₂)₂

Step 2: Define conversion

Molar Mass of Ba - 137.33 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Ba(NO₂)₂ - 137.33 + 2(14.01) + 4(16.00) = 229.35 g/mol

Step 3: Dimensional Analysis

$4.5 \cdot 10^{22} \ mc \ Ba(NO_2)_2(\frac{1 \ mol \ Ba(NO_2)_2}{6.022 \cdot 10^{23} \ mc \ Ba(NO_2)_2} )(\frac{229.35 \ g \ Ba(NO_2)_2}{1 \ mol \ Ba(NO_2)_2} )$

= 17.1384 g Ba(NO₂)₂

Step 4: Check

We are given 2 sig figs. Follow sig fig rules.

17.1384 g Ba(NO₂)₂ ≈ 17 g Ba(NO₂)₂

7 0

3 years ago

A race car is driven by a professional driver at 99 . What is this speed in and ? 1 mile = 1.61 kilometers 1 hour = 60 minutes E

slavikrds [6]

Answer: The speed is equivalent to 159.39 kilometers per hour or 2.65 kilometers per minute.

Explanation:

Given, The speed of a race car = 99 miles/ hour

To convert the speed into kilometers per hour and kilometers per minute

Since 1 mile = 1.61 kilometers

So, Speed of car = (99 ) x (1.61 )

= 159.39 kilometers per hour.

Also, 1 hour = 60 minutes

Then, Speed of car = (159.39) ÷60

= 2.6565≈2.65 kilometer per minute.

Hence, the speed is equivalent to 159.39 kilometers per hour or 2.65 kilometers per minute.

4 0

4 years ago

Atypicalaspirintabletcontains325mgofacetylsalicylic acid (HC9H7O4). Calculate the pH of a solution that is prepared by dissolvin

Sergio [31]

Answer:

$\boxed{2.65}$

Explanation:

1. Mass of acetylsalicylic acid (ASA)

$m = \text{2 tablets} \times \dfrac{\text{325 mg}}{\text{1 tablet}} = \text{750 mg}$

2. Moles of ASA

HC₉H₇O₄ =180.16 g/mol

$n = \text{750 mg} \times \dfrac{\text{1 mmol}}{\text{180.16 mg }} = \text{4.163 mmol}$

3. Concentration of ASA

$c = \dfrac{\text{4.163 mmol}}{\text{237 mL}} = \text{0.01757 mol/L}$

4. Set up an ICE table

$\begin{array}{ccccccc}\text{HA} & + & \text{H$_{2}$O}& \, \rightleftharpoons \, &\text{H$_{3}$O$^{+}$} & + &\text{A}^{-}\\0.01757 & & & &0 & & 0 \\-x & & & &+x & & +x \\0.01757-x & & & &x & & x \\\end{array}\\$

5. Solve for x

$K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}]\text{A}^{-}]} {\text{[HA]}} = 3.33 \times 10^{-4}\\\\\dfrac{x^{2}}{0.01757 - x} = 3.33 \times 10^{-4}\\\\\textbf{Check that }\mathbf{x \ll 0.01757}\\\\\dfrac{ 0.01757 }{3.33 \times 10^{-4}} = 53 < 400\\\\\text{The ratio is less than 400. We must solve a quadratic equation.}\\\\x^{2} = 3.33 \times 10^{-4}(0.01757 - x) \\\\x^{2} = 5.851 \times 10^{-6} - 3.33 \times 10^{-4}x\\\\x^{2} + 3.33 \times 10^{-4}x - 5.851 \times 10^{-6} = 0$

6. Solve the quadratic equation.

$a = 1; b = 3.33 \times 10^{-4}; c = -5.851 \times 10^{-6}$

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{Substituting values into the formula, we get}\\x = 0.002258\qquad x = -0.002591\\\text{We reject the negative value, so}\\x = 0.002258$

7. Calculate the pH

$\rm [H_{3}O^{+}]= x \, mol \cdot L^{-1} = 0.002258 \, mol \cdot L^{-1}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.002258} = \mathbf{2.65}\\\text{The pH of the solution is } \boxed{\textbf{2.65}}$

4 0

3 years ago

If the temperature of an ideal gas is raised from 100◦C to 200◦C, while the pressure remains constant, the volume 1. increases b

mel-nik [20]

Answer:

3. doubles

Explanation:

for an ideal gas behavior, the relationship between volume and temperature is given by Charles law

Charles law states that the volume of a given mass of gas is directly proportional to its temperature provided that pressure remains constant. Mathematically, this is represented as

V ∝ T

V=KT

K = V/T

where V is the volume of the gas

T is the Temperature

k represents the constant of proportionality

For initial and final conditions of a gas,

$\frac{V_{1} }{T_{1} }$ = $\frac{V_{2} }{T_{2} }$

where 1 and 2 represent initial and final conditions respectively

therefore, T₁ = 100 and T₂ = 200

$\frac{V_{1} }{100}$ = $\frac{V_{2} }{200}$

200 × V₁ = 100 × V₂

divide both sides by 100

2V₁ = V₂

final volume,V₂ = 2V₁

there the volume doubles

3 0

3 years ago

When the following oxidation-reduction occurs, what is the balanced reduction half-reaction after the electrons in both half rea

Lostsunrise [7]

Answer : The balanced reduction half-reaction is:

$3Cu^{2+}+6e^-\rightarrow 3Cu$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The given balanced redox reaction is :

$Al(s)+Cu^{2+}(aq)\rightarrow Al^{3+}(aq)+Cu(s)$

The half oxidation-reduction reactions are:

Oxidation reaction : $Al\rightarrow Al^{3+}+3e^-$

Reduction reaction : $Cu^{2+}+2e^-\rightarrow Cu$

In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 3 and then added both equation, we get the balanced redox reaction.

Oxidation reaction : $2Al\rightarrow 2Al^{3+}+6e^-$

Reduction reaction : $3Cu^{2+}+6e^-\rightarrow 3Cu$

The balanced redox reaction will be:

$2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s)$

Thus, the balanced reduction half-reaction is:

$3Cu^{2+}+6e^-\rightarrow 3Cu$

6 0

3 years ago