Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
Answer:
This is known as the coefficient factor
Explanation:The balanced equation makes it possible to convert information about one reactant or product to quantitative data about another element.
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.
The answer to this problem is 11.6m
Answer:
35 . 29%
Explanation:
no. of questions in test =34
no. of questions answered correctly =22
therefore, no. of questions answered incorrectly =34 - 22
=12
error percentage = 12/34 * 100
=35 .29 %
