Answer:
Sr(s) + C(s) + 3/2 O₂(g) → SrCO₃(s)
Explanation:
The standard enthalpy of formation (ΔH°f) is the energy involved in the formation of 1 mole of a substance from its elements in their most stable states. The chemical equation for the formation of SrCO₃(s) is the following.
Sr(s) + C(s) + 3/2 O₂(g) → SrCO₃(s)
Answer:
0.66g of water
Explanation:
Molar heat of vaporization of any substance is defined as the heat necessary to vaporize 1 mole of the substance.
If heat of vaporization of water is 40.79kJ/mol and you add 1.50kJ, the moles you vaporize are:
1.50kJ × (1mol / 40.79kJ) = 0.0368 moles of water.
As molar mass of water is 18.01g/mol, mass of water that can be vaporized are:
0.0368 moles × (18.01g / mol) = <em>0.66g of water</em>
Answer: The molecular formula will be 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 70.6 g
Mass of H = 5.9 g
Mass of O = 23.5 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = 
For H =
For O =
The ratio of C : H: O= 4: 4:1
Hence the empirical formula is 
The empirical weight of = 4(12)+4(1)+1(16)= 68g.
The molecular weight = 136 g/mole
Now we have to calculate the molecular formula.
The molecular formula will be=
A conversion factor is a fraction or a ratio representing a relationship of two different measurement values. To write 20% m/v to a conversion factor, we need to remember that a percent is a value that represents the amount of a part per 100 units of the whole. M/v in the given value represents that the percentage is by mass per volume. So, to write it as a conversion factor, we do as follows:
20% m/v = 20 mass units / 100 volume units = 1 mass units / 5 volume units
Usually units of this are in g per L. So, it is equivalent to 1 g / 5 L
Answer:
The coefficient of O2 is 11
Explanation:
Step 1:
The equation for the reaction:
FeS2 + O2 → SO2 + Fe2O3
Step 2:
Balancing the equation. The equation can be balance as follow:
FeS2 + O2 → SO2 + Fe2O3
There are 2 atoms of Fe on the right side and 1 atom on the left. It can be balance by putting 2 in front of FeS2 as shown below:
2FeS2 + O2 → SO2 + Fe2O3
There are 4 atoms of S on the left side and 1 atom on the right side. It can be balance by putting 4 in front of SO2 as shown below:
2FeS2 + O2 → 4SO2 + Fe2O3
Now, there are a total of 11 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 11/2 in front of O2 as shown below:
2FeS2 + 11/2O2 → 4SO2 + Fe2O3
Multiply through by 2 to clear the fraction as shown below:
4FeS2 + 11O2 → 8SO2 + 2Fe2O3
Now the equation is balanced.
The coefficient of O2 is 11
