Answer:
1. 0.338 moles of Fe
2. 0.700 moles of Fe
3. 48.3%
Explanation:
This is the reaction:
Fe₂O₃ + 3C → 2Fe + 3CO
We were told that we produce 18.9 g of Fe. Let's convert the mass to moles:
18.9 g . 1mol/ 55.85 g = 0.338 moles of Fe
Let's make a rule of three; ratio is 1:2.
1 mol of oxide can produce 2 moles of elemental iron
Then, 0.350 moles must produce (0.350 .2) / 1 = 0.700 moles of Fe
Let's determine the percent yield:
(Yield produced /Theoretical Yield) . 100 = 48.3 %
Answer:
The formulas that will be used are:
Situation one:
The moon will be experiencing gravitational force in the form of centripetal force, so we equate the two formulas.
Gravitational force = GMm /r²
Centripetal force = mv²/r
Equating,
GMm/r² = mv²/r
v² = GM/r
The first scenario will use the formula v² = GM/r
Situation 2:
The second situation will use the simple distance over time formula for velocity, where the distance will be the circumference and the time will be in seconds.
v = 2rπ/t
Explanation:
Answer:
Bottle, thermos, container, vessel
Explanation:
Answer : The value of for the reaction is, -521.6 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given main reaction is,
The intermediate balanced chemical reaction will be,
(1)
(2)
Now we are multiplying reaction 1 by 2 and reversing reaction 2 and then adding all the equations, we get :
(1)
(2)
The expression for enthalpy of change will be,
Thus, the value of for the reaction is, -521.6 kJ
Answer:
21.2 gm
Explanation:
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
butane is the hydrocarbon C4H10
in combustion, we react hydrocarbons with O2 to form CO2 and H2O
so
C4H10 + O2----------------> CO2 + H2O
BALANCE
2C4H10 + 1302--------> 8CO2 + 10 H2O
the molar mass of CO2 is 12 + 16X2 = 44
64.1 gm of CO2 is
64.1/44 = 1.46 MOLES OF CO2,
FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A
8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2
SO
THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES
THE MOLAR MASS OF BUTANE IS 58.12
0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm