Answer:
11419 J/g/ 11.419 KJ/g
Explanation:
H=MCQ
H=225×2.03×(-15-10)
H=225×2.03(25) Note; negative sign is of no use
H=11419J/g
Answer:
m = 4450 g
Explanation:
Given data:
Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)
Initial temperature = 23.0°C
Final temperature = 57.8°C
Specific heat capacity of water = 1 cal/g.°C
Mass of water in gram = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 57.8°C - 23.0°C
ΔT = 34.8°C
4450 cal = m × 1 cal/g.°C × 34.8°C
m = 4450 cal / 1 cal/g
m = 4450 g
Answer : 
is the oxidizing agent and Fe is the reducing agent.
Explanation :
Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.
Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.
The balanced redox reaction is :
The half oxidation-reduction reactions are:
Oxidation reaction : 
Reduction reaction : 
In order to balance the electrons, we multiply the oxidation reaction by 4 and reduction reaction by 3 then added both equation, we get the balanced redox reaction.
Oxidation reaction : 
Reduction reaction : 
In this reaction, 
is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and 
is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.
Thus, is the oxidizing agent and Fe is the reducing agent.
Answer:
Explanation:
The specific heats of water and steel are
Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium 
An energy balance can be written as
Replacing
Then, the temperature
Answer:
CH₄
Explanation:
To determine the empirical formula of the hydrocarbon, we need to follow a series of steps.
Step 1: Determine the mass of the compound
The mass of the compound is equal to the sum of the masses of the elements that form it.
m(CxHy) = mC + mH = 7.48 g + 2.52 g = 10.00 g
Step 2: Calculate the percent by mass of each element
%C = mC / mCxHy × 100% = 7.48 g / 10.00 g × 100% = 74.8%
%H = mH / mCxHy × 100% = 2.52 g / 10.00 g × 100% = 25.2%
Step 3: Divide each percentage by the atomic mass of the element
C: 74.8/12.01 = 6.23
H: 25.2/1.01 = 24.95
Step 4: Divide both numbers by the smallest one, i.e. 6.23
C: 6.23/6.23 = 1
H: 24.95/6.23 ≈ 4
The empirical formula of the hydrocarbon is CH₄.
