Answer: they are the same.
Explanation:
1) Sucrose is a compound with chemical formula C₁₂H₂₂O₁₁
2) That means that all molcules of sucrose will have the same kind of atoms in the same proportion, whic is to say same composition:
12 atoms of C: 22 atoms of H: 11 atoms of O, per each molecule of sucrose.
3) For this question you can rely in the definition of compound: a pure substance formed by the combination of two or more elements always in the same ratio (same composition).
4) That also implies, that all the molecules of sucrose have the same properties.
Explanation:
The given data is as follows.
Moles of propylene = 100 moles,
= 300 K
= 800 K,
,
of propylene = 100 J/mol
Now, we assume the following assumptions:
Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.
W = 

= 
= 5 MJ
Thus, we can conclude that a minimum of 5 MJ work is required without any friction.
Answer:
1. Percentage composition of: Na = 42%; P = 19.0%; O = 39%
2. Simplest formula of compound is PbO₂
3. (i) 2Cu(NO₃) ---> 2CuO + 2NO₂ + 3O₂
(ii) 2C₂H₆ + 7O₂ ---> 4CO₂ + 6H₂O
(iii) Mg₃N₂ + 6H₂O ---> 3Mg(OH)₂ + 2NH₃
4. 48 g of MG will react with 2 moles of Cl₂
5. 0.288 g of SO2 will be produced from the combustion of 0.331 g P₄S₃ in excess O₂
6. 12.8 g of nitric oxide can be produced from the reaction of 8.00 g NH₃ with 17.0 g O₂
7. The stock acid solution should be diluted to 6000 mL or 6.0 L
Explanation:
The full explanation is found in the attachments below
The more numbers after the decimal point there are, the more precise the instrument which recorded it is. For example, if one instrument during seismic activity records that the magnitude of the earthquake was 2.3, and another instrument recorded that it was 2.3645, the second instrument would have shown to be more precise.
<u>Answer:</u> The enthalpy of the reaction for the production of
is coming out to be -74.9 kJ
<u>Explanation:</u>
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^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-74.9%29%29%5D-%5B1%5Ctimes%200%29%2B%282%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-74.9kJ)
Hence, the enthalpy of the reaction for the production of
is coming out to be -74.9 kJ