Answer:
The enthalpy of the reaction is coming out to be -380.16 kJ.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O%29%7D%29%2B%282%20mol%5Ctimes%5CDelta%20H_f_%7B%28H_2O%29%7D%20%29%5D-%5B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2H_4%29%7D%29%2B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O_4%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%2081.6%20kJ%2Fmol%29%2B2%20mol%5Ctimes%20-241.8%20kJ%2Fmol%29%5D-%5B%281%20mol%5Ctimes%20%2850.6%20kJ%2Fmol%29%29%2B%281%20mol%5Ctimes%20%289.16%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-380.16%20kJ)
Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.
Answer:
B. How many of each atom are present in the compound
D. the simplified ratio of atoms in relation to each other
Explanation:
In a chemical formula, chemical elements or atoms are represented by a chemical symbol for example Fe for iron and Na for sodium, and the number of each atom is represented by a subscript such as CO2, where 2 is a subscript representing 2 atoms of oxygen.
A subscript represents the number of each atom in the compound and the simplified ratio of atoms in relation to each other. The simplified ratio of atoms in relation to each other means subscript shows the contribution of both the atoms in the compound, for example: N2 + 3H2 => 2NH3, it means the subscript showing the ratio or proportionate of atoms that is 2:2 for both nitrogen and hydrogen.
The subscript is always written below and to the right of the chemical symbol.
Hence, the correct answer is "B. How many of each atom are present in the compound and D. the simplified ratio of atoms in relation to each other"
Answer:
Empirical formula is CaSO₄.
Explanation:
Given data:
Percentage of calcium =29.44%
Percentage of sulfur = 23.55%
Percentage of oxygen = 47.01%
Empirical formula = ?
Solution:
Number of gram atoms of Ca = 29.44 / 40 = 0.74
Number of gram atoms of S = 23.55 / 32 = 0.74
Number of gram atoms of O = 47.01 / 16 = 3
Atomic ratio:
Ca : S : O
0.74/0.74 : 0.74/0.74 : 3/0.74
1 : 1 : 4
Ca : S : O = 1 : 1 : 4
Empirical formula is CaSO₄.
The balanced chemical reaction is given as follows:
<span>2 KClO3(s) → 2 KCl(s) + 3 O2(g)
The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:
</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
