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vladimir2022 [97]
2 years ago
11

A nucleotide is composed of a(n) _____. phosphate group, a nitrogen-containing base, and a five-carbon sugar amino group, a nitr

ogen-containing base, and a five-carbon sugar sulfhydryl group, a nitrogen-containing base, and a five-carbon sugar glycerol, a nitrogen-containing base, and a five-carbon sugar phosphate group, a nitrogen-containing base, and a hydrocarbon
Chemistry
1 answer:
lapo4ka [179]2 years ago
5 0

A nucleotide is composed of a phosphate group, a nitrogen-containing base, and a five-carbon sugar amino group. A nucleotide is composed of a phosphate group, a nitrogen-containing base, and a five-carbon sugar amino group. A nucleotide is the building block or structural component of DNA and RNA. It consists of a base , that is one from adenine, thymine, guanine, and cytosine. and a molecule of sugar and one of phosphoric acid.


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What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio
Veseljchak [2.6K]

Answer:

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol

Mass of methane burnt = 530 g

Moles of methane burnt = \frac{530 g}{16 g/mol}=33.125 mol

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ

-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

7 0
2 years ago
A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration
Sliva [168]

Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

Explanation:

The volume of vessel used= 100 ml

Initial moles of NO= \frac{2.60}{10^2} moles

Initial moles of H2= \frac{1.30}{10^2} moles

Concentration of NO at equilibrium= 0.161M

Concentration(in M)=\frac{moles}{volume(in litre)}

Moles of NO at equilibrium= 0.161(\frac{100}{1000})

                                            =\frac{1.61}{10^2} moles

               

                    2H2 (g)        +    2NO(g) <—>    2H2O (g) +    N2 (g)

<u>Initial</u>          :1.3*10^-2          2.6*10^-2                0                   0        moles

<u>Equilibrium</u>:1.3*10^-2 - x     2.6*10^-2-x              x                   x/2     moles

∴\frac{2.60}{10^2}-x=\frac{1.61}{10^2}

⇒x=\frac{0.99}{10^2}

Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)

<u>Equilibrium</u>:0.31*10^-2      1.61*10^-2          0.99*10^-2        0.495*10^-2  moles

⇒Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2}  (0.1)

⇒Kc=19.4760

3 0
2 years ago
A piece of silver releases 202.8 J of heat while cooling 65.0 °C. What is the mass of the sample? Silver has a specific heat of
KiRa [710]

Answer: 13 grams

Explanation:

The quantity of heat energy (Q) released from a heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = 202.8 Joules

Mass of silver = ?

C = 0.240 J/g °C.

Φ = 65°C

Then, Q = MCΦ

202.8J = M x 0.240 J/g °C x 65°C

202.8J = M x 15.6 J/g

M = (202.8J / 15.6 J/g)

M = 13 g

Thus, the mass of silver is 13 grams

7 0
2 years ago
3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO42−(aq) + 8 H2O(l) ΔG∘ = +87 kJ/molGiven the standard reduction potenti
Mrrafil [7]

Answer:

The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V

Explanation:

<u>Step 1</u>: Data given

3 Ni^2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO4^2−(aq) + 8 H2O(l) ΔG∘ = +87000 J/mol

Ni2+(aq) + 2 e− → Ni(s)    E∘red = -0.28 V

<u>Step 2:</u> The half reactions:

Cathode:  Ni2+(aq) + 2 e− → Ni(s)    E° = -0.28 V

Anode: CrO4^2-(aq) + 4H2O(l) +3e- → Cr(OH)3(s) + 5OH- (aq)   E°= unknown

<u>Step 3:</u> Calculate E°cell

ΔG° = -n*F*E°cell

⇒ with ΔG° = the gibbs free energy

⇒ n = the number of electrons in the net reaction = 6

⇒ F = the Faraday constant = 96485 C

⇒ E°cell= the standard cell potential

<u>Step 4:</u> Calculate E°(Cr6+/Cr3+

E°cell= ΔG°/(-n*F)

E°cell = 87000 /(-6*96485)

E°cell = -0.150 V

E°cell = E°(Ni2+/Ni) - E°(Cr6+/Cr3+)

E°(Cr6+/Cr3+) = -0.13V

The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V

7 0
2 years ago
A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it
patriot [66]

Answer:

P1 = 2.5ATM

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15K

P1 = ?

P2 = 2ATM

applying combined gas equation,

P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

Solving for P1

P1 = P2*V2*T1 / V1*T2

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

P1 = 21634.2 / 8628.2

P1 = 2.5ATM

The initial pressure was 2.5ATM

3 0
2 years ago
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