Answer:
C. The reaction can be broken down and performed in steps
Explanation:
Hess's Law of Constant Heat Summation states that irrespective of the number of steps followed in a reaction, the total enthalpy change for the reaction is the sum of all enthalpy changes corresponding to all the steps in the overall reaction. The implication of this law is that the change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states of the system.
To obtain MgO safely without exposing magnesium to flame, the reaction sequence shown in the image attached may be carried out. Since the enthalpy of the overall reaction is independent of the pathway between the initial and final states of the system, the sum of the enthalpy of each step yields the enthalpy of formation of MgO.
Answer:
8/3 hours
Explanation:
12.5/100=1/8, 1/8*2=2/8=1/4 1/4*2=1/2 1/2*2 = 1 (whole)
we multiplyed 1/8 by two three times to get 1 so 8/3 hours is the answer
by the way i'm in 6th grade
starting with q.1 its the 2nd and 7th day
q.2 the 12th
q.3 is 8 days
q.4 its the second 6 days
q.5 9 days
q.6 4 days
q.7 5 of the days
Answer:
Theoretical yield of HI is 512 g.
The percent yield for this reaction is 25%.
Explanation:
Moles of hydrogen gas = 3.0 moles
Moles of iodine gas = 2.0 moles
According to reaction 1 mol of hydrogen gas reacts with 1 mol of iodine gas.
Then 3.0 moles of hydrogen gas reacts with 3.0 mol of iodine gas. But there are 2.0 moles of iodine gas. Hence,Iodine is a limiting reagent. The production of HI will depend upon iodine gas moles.
According to reaction , 1 mol of iodine gas gives 2 moles of HI.
Then 2 moles of iodine gas will give:
of HI
Theoretically we will get 4 moles of HI.
Theoretical yield of HI = 4 mol × 128 g/mol= 512 g
Experimental yield of HI = 1.0 mol
= 1 mol × 128 g/mol= 128 g
The percent yield for this reaction is 25%.
Explanation:
Let compound X be 100g.
Molar mass of carbon = 12g/mol
Moles of C in compound X = 54.54/12 = 4.54mol.
Molar mass of hydrogen = 1g/mol
Moles of H in compound X = 9.09/1 = 9.09mol.
Molar mass of oxygen = 16g/mol
Moles of O in compound X = 36.37/16 = 2.27mol
Ratio of C:H:O = 4.54:9.09:2.27 = 2:4:1.
Hence, the empirical formula of X is C2H4O.
