Answer:
The number ratio is 4:7
Explanation:
Step 1: Data given
Compound 1 has 50.48 % oxygen
Compound 2 has 36.81 % oxygen
Molar mass oxygen = 16 g/mol
Molar mass manganese = 54.94 g/mol
Step 2: Calculate % manganes
Compound 1: 100 - 50.48 = 49.52 %
Compound 2: 100 - 36.81 = 63.19 %
Step 3: Calculate mass
Suppose mass of compounds = 100 grams
Compound 1:
50.48 % O = 50.48 grams
49.52 % Mn = 49.52 grams
Compound 2:
36.81 % O = 36.81 grams
63.19 % Mn = 63.19 grams
Step 4: Calculate moles
Compound 1
Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles
Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles
Compound 2
Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles
Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles
Step 5: calculate mol ratio
We will divide by the smallest amount of moles
Compound 1
O: 3.155/0.9013 = 3.5
Mn: 0.9013 / 0.9013 = 1
Mn2O7
Compound 2
O: 2.301 / 1.150 = 2
Mn: 1.150 / 1.150 = 1
MnO2
The number ratio is 2:3.5 or 4:7
When a water vapor condenses, heat is being released from the process. This heat is called latent heat of vaporization since the phase change happens without any change in the temperature. This value is constant per mole of a substance as a function of pressure and temperature. For this problem, we are given the heat of vaporization at a certain T and P. We use this value to calculate the total heat released from the process. We calculate as follows:
Total heat released: 32.4 g ( 1 mol / 18.02 g ) (40.67 kJ / mol) = 73.12 kJ
Therefore, 73.12 kJ of heat is released from the condensation of 32.4 g of water vapor.
Answer is: the ratio of the effusion rate is 1.59 : 1.
1) rate of effusion of carbon monoxide gas = 1/√M(CO).
rate of effusion of carbon monoxide gas = 1/√28.
rate of effusion of carbon monoxide gas = 0.189.
2) rate of effusion of chlorine = 1/√M(Cl₂).
rate of effusion of chlorine = 1/√70.9.
rate of effusion of chlorine = 0.119.
rate of effusion of carbon monoxide : rate of effusion of chlorine =
= 0.189 : 0.119 / ÷0.119.
rate of effusion of carbon monoxide : rate of effusion of chlorine = 1.59 : 1.
We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams