All categories

3 years ago

Calculate the relative density of carbon monoxide and carbon dioxide with respect to air.

1 answer:

3 0

Density is proportional to molar mass, assuming pressure and temperature remain constant. Therefore, since CO has a molar mass of 28 and CO2 has a molar mass of 44:
The relative density of CO vs air is 28/29 = 0.9655.
The relative density of CO2 vs air is 44/29 = 1.517.

You might be interested in

What will occur if the velocity of the gas molecules colliding with the sides of the container increases?

yanalaym [24]

The pressure of the gas will increases

6 0

3 years ago

Read 2 more answers

What are chemical formulaes? give your own response!!

lbvjy [14]

Answer:

is a way of presenting information about the chemical proportions of atoms

Explanation:

8 0

3 years ago

How is nitric acid classified?

Darina [25.2K]

It's classified as an acid

4 0

3 years ago

A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and

Lunna [17]

The pH of the unknown solution is 3.07.

Explanation:

1.Find the cell potential as a function of pH

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

where

R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0 lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

2. Find the Q value

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

E cell= 2.303 × 8.314 CK mol (inverse) × 298.15

K × 2pH /2×96 485 C⋅mol

( inverse)

E cell= 0.0592 V × pH

3. Finding the pH value

E cell= 0.0592 V × pH

pH = E cell/ 0.0592 V= 0.182V/ 0.0592V

pH=3.07

The pH of the unknown solution is 3.07.

7 0

3 years ago

11C. Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solut

Llana [10]

For the reactants,

The oxidation number of hydrogen = +1
The oxidation number of oxygen = -2
The oxidation number of arsenic = +5
The oxidation number of carbon = +3

For the products,

The oxidation number of hydrogen = +1
The oxidation number of oxygen = -2
The oxidation number of arsenic = +3
The oxidation number of carbon = +4

Here, arsenic (+5 to +3) and carbon (+3 to +4) are the only oxidation numbers changing.

Note that an increase in oxidation number means electrons are lost. Thus oxidation is occurring, and a decrease in oxidation number means electrons are being gained, and thus reduction is occurring.

Also, the compound that contains the element being oxidized is the reducing agent, and the compound that contains the element being reduced is the oxidizing agent.

So, the answers are:

name of the element oxidized: Carbon

name of the element reduced: Arsenic

formula of the oxidizing agent: $\text{H}_{3}\text{AsO}_{4}$

formula of the reducing agent: $\text{H}_{2}\text{C}_{2}\text{O}_{4}$

6 0

2 years ago