As less heat will escape out of the house. Heat escapes out the loft, window or floors
Answer:
The answer to your question is C₄H₅N₂O
Explanation:
Process
1.- Calculate the percent of oxygen in the sample
Percent of oxygen = 100 - 49.49 - 5.15 - 28.87
Percent of oxygen = 16.49 %
2.- Write the percents as grams
C = 49.49 g
H = 5.15 g
N = 28.87 g
O = 16.47 g
3.- Convert the grams to moles
C 12 g ------------------- 1 mol
49.49 g ---------------- x
x = (49.49 x 1) 12
x = 4.12 moles
H 1 g ------------------- 1 mol
5.15 g ---------------- x
x = (5.15 x 1)/ 1
x = 5.15 moles
N 14 g --------------- 1 mol
28.87 g ---------- x
x = (28.87 x 1) / 14
x = 2 mol
O 16 g ---------------- 1 mol
16.49 g ----------- x
x = (16.49 x 1) / 16
x = 1.03 moles
4.- Divide by the lowest number of moles
C 4.12 / 1.03 = 4
H 5.15 / 1.03 = 5
N 2 / 1.03 = 1.9 ≈ 2
O 1.03 / 1.03 = 1
5.- Write the empirical formula
C₄H₅N₂O
The balanced reaction is
3Na3PO4 + 2CuSO4 ------> 3Na2SO4 + Cu3(PO4)2
To balance this reaction of double displacement, we see first that this reaction maintain the valence numbers of every atom.
Then, to have the same value of Na in the two sides of the reaction we multiply for the number of the other side. So,
(Na3PO4)x 2
(Na2SO4)x3
As we can see either, we need to balance PO4 cause there are two molecules of this in the reactant side, so we have two molecules of PO4 in the product either.
Then we get
3Na3PO4 + 2CuSO4 ------> 3Na2SO4 + Cu3(PO4)2
To probe that balance was correct, you can verify that the charges are exactly the opposite.
<u>Answer:</u> The enthalpy of the formation of 
is coming out to be -410.8 kJ/mol.Z
<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%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%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_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\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%28C_2H_2%28g%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%285%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:
![81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol](https://tex.z-dn.net/?f=81.1%3D%5B%281%5Ctimes%20%28226.7%29%7D%29%2B%284%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%285%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%3D-410.8kJ%2Fmol)
Hence, the enthalpy of the formation of is coming out to be -410.8 kJ/mol.
Answer:
double replacement
Explanation:
The reaction equation is given as:
Pb(NO₃)₂ + 2KI → PbI₂ + KNO₃
This reaction is a double displacement reaction.
In this form of reaction, there is an actual exchange of partners to form new compounds.
AB + CD → AD + CD
One of the following condition serves as driving force for the reaction:
- formation of an insoluble compound
- formation of water or any other non-ionizing compound
- liberation of a gaseous product.
