1) Balance of mass
CxHyOz + O2 ---> CO2 + H20
0.880 g + X = 1.760g + 0.720 g
=> X = 1.760g + 0.720g - 0.880g = 1.6 g of O2
2) C in CO2:
molar mass of CO2 = 44 g/mol
=> 12 g C / 44 g CO2 * 1.760 g CO2 = 0.480 g C
3) H in H2O
molar mass of H2O = 18.0 g/mol
2 g H / 18.0 g H2O * 0.720 g H2O = 0.080 g H
4) O
4a) O in CO2: 32gO / 44 g CO2 * 1.760 g CO2 = 1.280 g O
4b) O in H2O: 16gO / 18 gH2O * 0.720 g H2O = 0.640 g O
4c) O in CxHyOz = 1.280 g + 0.64 g - 1.6 g = 0.320 g O
5) Convert grams into moles
C: 0.480 g / 12 g/mol = 0.04 mol
H: 0.08 g / 1 g/mol = 0.08 mol
O: 0.32 g / 16 g/mol = 0.02 mol
6) Divide by the smallest number
C: 0.04 / 0.02 = 2
H: 0.08 / 0.02 = 4
O: 0.02 / 0.02 = 1
7) Use those numbers as subscripts for the empirical formula:
C2 H4 O
Answer: C2H4O
Answer:
Jim's car travels: 12.04 meters per second.
Tara's car travels 12 meters per second.
Explanation:
Tara's car travels the fastest.
Answer : The value of the equilibrium constant is,
Explanation :
The given main reaction is:
The intermediate reactions are:
;
;
;
Now we are adding all the equation, we get:
...........(1)
The equilibrium constant expression will be:
Now we are dividing equation 1 by 2, we get:
...........(1)
The equilibrium constant expression will be:
Now put all the given values in this expression, we get:
Thus, the value of the equilibrium constant is,
Answer:
CaC₂(s) + 2H₂O(l) --> C₂H₂(g) + Ca(OH)₂(aq)
Explanation:
We are given;
The unbalanced equation;
CaC₂(s) + 2H₂O(l) --> C₂H₂(g) + Ca(OH)₂(aq)
We are required to balance the equation;
- We need to know that balancing of chemical equations involves putting the appropriate coefficients on reactants and products to ensure equal number of atoms of each element on both sides of the equation.
- Balancing of chemical equations is a try and error method of making sure that the law of conversation of mass in chemical equations is obeyed.
- In our case, the appropriate coefficients will be, 1, 2, 1, 1
Therefore;
The balanced equation will be;
CaC₂(s) + 2H₂O(l) --> C₂H₂(g) + Ca(OH)₂(aq)