Convert 14 mm to its equivalent measurement in m.

Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t

ratelena [41]

Answer:

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

$2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}$

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

$\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol$

$\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol$

$\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol$

$\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})$

$\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)$

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

The equilibrium constant determined from the partial pressure denoted as $K_p$ can be expressed as :

$K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}$

$K_p = \dfrac{1}{ (22.20)}$

$K_p$ = 0.045

$\Delta G = \Delta G^0 _{rxn} + RT \ lnK$

where;

R = gas constant = 8.314 × 10⁻³ kJ

$\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)$

$\Delta G =207.6 + 2.4788191 \times \ ln(0.045)$

$\Delta G =207.6+ (-7.687048037)$

$\Delta G =$ 199.912952 kJ

ΔG ≅ 199.91 kJ

7 0

4 years ago

Hemoglobin has a much greater affinity for carbon monoxide than oxygen. which principle explains why a hyperbaric chamber (conta

babunello [35]

Hemoglobin has a much greater affinity for carbon monoxide than oxygen. In a hyperbaric chamber (containing high levels of oxygen) can treat carbon monoxide poisoning, by displacing carbon monoxide from Hemoglobin competitively.

Hemoglobin has a much greater affinity for carbon monoxide than oxygen. This is because, a coordinate bond is formed with Carbon monoxide and Haem structure of the hemoglobin.

Carbon monoxide with Hemoglobin is called as Carboxy haemoglobin.

Presence of oxygen displaces the Carbon monoxide with Hemoglobin that is formed due to poisoning.

Hyperbaric chamber is a chamber which contains pure oxygen in a chamber. The atmospheric pressure is kept about three to four times than the normal, such that the replacement of Carbon monoxide from Haem can occur as fast as possible since this reduces the half life of the Carboxy haemoglobin.

It is advisable not to treat Carbon monoxide poisoning yourself.

Hyperbaric oxygen is used to treat the following conditions as well:

Infections
Wounds
Air bubble is blood

Learn more about Carbon Monoxide here, brainly.com/question/11313918

#SPJ4

3 0

2 years ago

If 27.50 mL of 0.120 M NaOH neutralizes 0.248 g of HAA, what is the molar mass of unknown amino acid

tino4ka555 [31]

Answer:

75.15 g/mol

Explanation:

First, let us look at the equation of reaction;

$NaOH + HAA --> NaAA + H_2O$

From the balanced equation of reaction, 1 mole of NaOH is required to completely neutralize 1 mole of HAA.

Recall that: mole = molarity x volume.

Therefore, 27.50 mL, 0.120 M NaOH = 0.0275 x 0.120 = 0.0033 moles

0.0033 mole of NaOH will therefore requires 0.0033 moles of HAA for complete neutralization.

In order to find the molar mass of the unknown amino acid, recall that:

<em>mole = mass/molar mass</em>, hence, <em>molar mass = mass/mole</em>.

Therefore, molar mass of HAA = 0.248/0.0033 = 75.15 g/mol

6 0

3 years ago

HELPP PLEASEEEEEEE ITS PROBS EASY FOR YOU GUYS

hichkok12 [17]

Answer:

ik that energy makes us movie and that it powers lights and electronics

kinetic energy ,chemical energy ,potential energy ,and thermal energy

the law of conservation of energy states that the total energy of an isolated system remains constant

Explanation:

hope it helpss

5 0

4 years ago

Read 2 more answers

Calculate the volume occupied by 0.845 mole of nitrogen gas, the pressure, and the temperature. Find the volume.

NeTakaya

Answer:

V = 15.9512 dm³

Explanation:

Given data:

Pressure = P = 1.37 atm

Temperature = T= 315 K

Number of moles of nitrogen= n = 0.845 mol

Volume = V = ?

Formula:

PV = nRT

Now we will put the values in equation.

V = nRT/ P

V = ( 0.845 mol× 0.0821 dm³.atm.K⁻¹.mol⁻¹ × 315 K) / 1.37 atm

V = 21.853 dm³. atm/ 1.37 atm

V = 15.9512 dm³

5 0

4 years ago

Read 2 more answers