Answer : The oxidizing element is N and reducing element is O.
is act as an oxidizing agent as well as reducing agent.
Explanation :
An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.
Reducing agent is the agent which has ability to reduce other or lower in oxidation number.
The given reaction is :
act as an oxidizing agent.
The oxidation number of N in is calculated as:
(+1)+(x)+3(-2) = 0
x = +5
And the oxidation number of N in is calculated as:
(+1)+(x)+2(-2) = 0
x = +3
From the oxidation number method, we conclude that the oxidation number reduced this means itself get reduced to and it can act as an oxidizing agent.
act as a reducing agent.
The oxidation number of O in is calculated as:
(+1)+(+5)+3(x) = 0
x = -2
The oxidation number of O in is Zero (o).
Now, we conclude that the oxidation number increases this means itself get oxidized to and it can act as reducing agent.
Answer:
By designing new technology that would prove new information
Explanation:
Answer:
22.9 Liters CO(g) needed
Explanation:
2CO(g) + O₂(g) => 2CO₂(g)
? Liters 32.65g
= 32.65g/32g/mol
= 1.02 moles O₂
Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)
∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)
Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.
∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed
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*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).
Answer:
Dispersion forces
Relative molecular mass
Explanation:
Alkanes experience only dispersion forces. Dispersion forces increase with increasevin the relative molecular mass of the compounds. Hence a higher relative molecular mass implies greater dispersion forces and a greater boiling point.