Equilibrium is reached when the net reaction is zero. The reaction is consistent, when reactants and products are neither excessively consumed, nor produced.
The rate of the forward reaction is equal to the rate of the reverse reaction.
AH1 = m * c1 * AT1 calculate this for ice (-25C to 0C) AH2 = AHfus(1 mole)=6.01 kJ = 6010 J AH3 = m *c3 * AT3 calculat this for water (0C to 100C) AH4 = AHvap(1mole)=40.67 kJ = 40670 J AH5= m * c5 * AT5 calculate this for steam (100C to 125C)
Sum ---- AH1+AH2+AH3+AH4+AH5
Data m=18g (1mole water)
c1=specific heat ice= 2.09 J/g K c3=specific heat water= 4.18 J/g K c5=specific heat steam= 1.84 J/g K
AT = (Tend - Tinitial) as this is a difference between temperatures it doesn't matter the units Celsius or Kelvin. Kelvin (K)=Celsius (C)+273.15
AT1 = 0C - (-25C)= 25C= 273.15K - 248.15K= 25K AT3= 100C - 0C = 100C= 100K AT5= 125C - 100C= 25C=25K
Answer:- 335 kcal of heat energy is produced.
Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:
From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.
We could solve this using dimensional analysis as:
= 1401.5 kJ
Now, let's convert kJ to kcal.
We know that, 1kcal = 4.184kJ
So, 
= 335 kcal
Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.
Answer:
This is a single replacement reaction because I replaces Br.
We know that:
mass = density x volume
The volume of the total mixture is 0.1 m³
Let the first liquid be A and the second be B
Mass Total = Mass A + Mass B
800 x 0.1 = 1500Va + 500(0.1 - Va)
30 = 1000Va
Va = 0.03 m³
Vb = 0.1 - 0.03 = 0.07 m³
