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 
for HCN (g) in the reaction is 135.1 kJ/mol.
<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. 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\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28NH_3%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H_f_%7B%28O_2%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28CH_4%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ](https://tex.z-dn.net/?f=-870.8%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%28-80.3%29%29%2B%283%5Ctimes%20%280%29%29%2B%282%5Ctimes%20%28-74.6%29%29%5D%5C%5C%5C%5C%5CDelta%20H_f_%7B%28HCN%29%7D%3D135.1kJ)
Hence, the for HCN (g) in the reaction is 135.1 kJ/mol.
Hey there!:
Molarity of NaOH = 6 M or 6 mol/L
Volume of NaOH = 40 mL
Therefore , number of moles of NaOH:
40 mL =( 6 mol / 1000 mL)*40=
6/ 1000 * 40 => 0.24 moles of NaOH
Hope this helps!
Answer:
See Explanation
Explanation:
Both oxygen and dinitrogen (1) oxide are known to rekindle a glowing splint. However, oxygen is an odorless gas while dinitrogen (1) oxide has a faint smell and has an effect on the central nervous system.
Also, the combustion of dinitrogen (1) oxide produces brown fumes of nitrogen IV oxide.
Finally, when dinitrogen (1) oxide and oxgen are both heated with copper, residual nitrogen gas is left with dinitrogen (1) oxide while no residual gas is left with oxygen.
