What type of bond occurs between atoms having nearly equal attraction for the electrons?

We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh

k0ka [10]

Answer:

The ideal molar volume is $\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The Z factor is $Z = 0.09997$

The real molar volume is $\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

Explanation:

From the question we are told that

The pressure is $P = 23 \ bar = 23 *10^5 Pa$

The temperature is $T = 30 ^ oC = 303 \ K$

According to the ideal gas equation we have that

$PV = nRT$

=> $\frac{V}{n}=V_z= \frac{RT}{P}$

Where $\frac{V}{n }$ is the molar volume and R is the gas constant with value

$R = 8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}$

substituting values

$\frac{V}{n} =V_z= \frac{ 8.314 * 303}{23 *10^{5}}$

$\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The compressibility factor of the gas is mathematically represented as

$Z = \frac{P * V_z}{RT}$

substituting values

$Z = \frac{23 *10^{5} * 0.001095}{8.314 * 303}$

$Z = 0.09997$

Now the real molar volume is evaluated as

$\frac{V_r}{n} = V_k= \frac{Z * RT }{P}$

substituting values

$\frac{V_r}{n} = V_k= \frac{0.09997 * 8.314 * 303}{23 *10^{5}}$

$\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

8 0

3 years ago

A 0.055 mol sample of formaldehyde vapor, CH2O, was placed in a heated 500 mL vessel and some of it decomposed. The reaction is

Elis [28]

Answer:

Kc for this reaction is 0.06825

Explanation:

Step 1: Data given

Number of moles formaldehyde CH2O = 0.055 moles

Volume = 500 mL = 0.500 L

At equilibrium, the CH2O(g) concentration = 0.051 mol

Step 2: The balanced equation

CH2O <=> H2 + CO

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CH2O] = 0.055 moles . 0.500 L

[CH2O] = 0.11 M

[H2] = 0M

[CO] = 0M

Step 4: The concentration at the equilibrium

[CH2O] = 0.11 - X M = 0.051 M

[H2] = XM

[CO] = XM

[CH2O] = 0.11 - X M = 0.051 M

X = 0.11 - 0.051 = 0.059

[H2] = XM = 0.059 M

[CO] = XM = 0.059 M

Step 5: Calculate Kc

Kc = [H2][CO]/[CHO]

Kc = (0.059 * 0.059) / 0.051

Kc = 0.06825

Kc for this reaction is 0.06825

7 0

3 years ago

Carbon dioxide is a compound of the elements carbon and oxygen (CO2). What happened to the carbon and oxygen to create this comp

dexar [7]

Answer:

D. they exchanged electrons

6 0

2 years ago

What mass of C6H8O7 should be used every 7.0 X 10^2mg NaHCO3

Savatey [412]

Mass C₆H₈O₇ : 0.531484 g

<h3>Further explanation</h3>

Reaction

3NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + 3 H₂O (l) + Na₃C₆H₅O₇ (aq)

MW NaHCO₃ : 84 g/mol

mass NaHCO₃ : 7.10² mg=0.7 g

mol NaHCO₃ :

$\tt mol=\dfrac{0.7}{84}=0.0083$

mol C₆H₈O₇ :

$\tt \dfrac{1}{3}\times 0.0083=0.00277$

MW C₆H₈O₇ : 192 g/mol

mass C₆H₈O₇ :

$\tt mass=0.00277\times 192=0.53184$

4 0

3 years ago

The aldol condensation involves nucleophilic attack of an enolate to a carbonyl. Use the pop-up menus to identify the total numb

Sindrei [870]

Aldol condensation involves the reaction of an acid or base with a carbonyl group producing a nucleophile that attacks another carbonyl compound to yield a β-hydroxyaldehyde or β-hydroxyketone compound.

<h3>What is aldol condensation?</h3>

The aldol condensation is a reaction in organic chemistry in which there is a reaction between an acid or base and a carbonyl group which then serves as the nucleophile that attcks a second carbonyl to yield a β-hydroxyaldehyde or β-hydroxyketone compound.

The aldol condensation may be acid catalysed or base catlysed. The question is incomplete hence the complete mechanimsms can not be decuced.

Learn more about aldol condensation: brainly.com/question/9415260

3 0

2 years ago