What is the relationship between a codon and an anticodon?

exactly 1mol of n2o4 is placed in an empty 1 l container. if at equilibrium n2o4 is dissociated 20%, what is the value of equili

egoroff_w [7]

Answer:

K = 0.2

Explanation:

Based on the chemical dissociation of N₂O₄:

N₂O₄ ⇄ 2NO₂

The equilibrium constant, K, of the reaction is:

K = [NO₂]² / [N₂O₄]

Now, if 20% of N₂O₄ is dissociated, 80% remains as N₂O₄ = 0.8mol/L = 0.8M

as 20% is dissociated, 0.2moles of N₂O₄ were dissociated and 0.2*2 = 0.4mol/L of NO₂ are produced.

Replacing in K:

K = [0.4M]² / [0.8M]

5 0

3 years ago

Characteristics of a substance that can be measured or observed by changing the identity of the material is a

algol [13]

C cause ur changing the identity of the material

4 0

3 years ago

Bicarbonate concentrate mixers may have a which are replaced on a routine basic.

Sidana [21]

Answer: True the bicarbonate mixture can help save time and few routine.

Explanation:

For the purpose of making dialysate for hemodialysis patient therapies a bicarbonate mixing and delivering systems designed to prepare a liquid sodium bicarbonate formulation comes in handy.

Certain systems like the SDS unit also allow for the transfer and distribution of acid concentrate solutions. We also provide stand-alone acid concentrate delivery systems using a variety of holding tanks and delivery methods.

A challenge for hemodialysis providers is to properly provide bicarbonate solution in a cost effective manner. Preparation and disinfection can be time-consuming and labor intensive.

Bicarbonate however can corrode certain metals and painted surfaces leaving your preparation area encrusted and grimy.

Furthermore, if not mixed properly, bicarbonate can negatively affect the dialysate solution.

The answer to the above is true the bicarbonate mixture can help save time and few routine.

3 0

3 years ago

Convert: 65 kg to mg 65,000 mg O 65,000,000 mg O 65 mg

makkiz [27]

Answer:

kg, mg. 65.00, 65,000,000. 65.01, 65,010,000. 65.02, 65,020,000. 65.03, 65,030,000. 65.04, 65,040,000. 65.05, 65,050,000. 65.06, 65,060,000.

Explanation:

3 0

3 years ago

Read 2 more answers

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) Minimum mass of air

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago