Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
<u><em>Answer:</em></u>



<u><em>Explanation:</em></u>
<u>Part 1: Solving for m</u>
<u>We are given that:</u>
E = mc²
To solve for m, we will need to isolate the m on one side of the equation
This means that we will simply divide both sides by c²

<u>Part 2: Solving for c</u>
<u>We are given that:</u>
E = mc²
To solve for c, we will need to isolate the m on one side of the equation
This means that first we will divide both sides by m and then take square root for both sides to get the value of c

<u>Part 3: Solving for E</u>
<u>We are given that:</u>
m = 80 and c = 0.4
<u>To get the value of E, we will simply substitute in the given equation: </u>
E = mc²
E = (80) × (0.4)²
E = 12.8 J
Hope this helps :)
<u>Answer:</u> For the given equation, only iron has the value of
equal to 0 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%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_{(Fe(s))})+(3\times \Delta H^o_f_{(CO_2(g))})]-[(3\times \Delta H^o_f_{(CO(g))})+(2\times \Delta H^o_f_{(Fe_2O_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe%28s%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%5D-%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe_2O_3%28s%29%29%7D%29%5D)
The enthalpy of formation for the substances present in their elemental state is taken as 0.
Here, iron is present in its elemental state which is solid.
Hence, for the given equation, only iron has the value of
equal to 0 kJ.
Answer:
Double bond
Chemical compounds with double bonds.
Ethylene Carbon-carbon double bond.
Acetone Carbon-oxygen double bond.
Dimethyl sulfoxide Sulfur-oxygen double bond.
Diazene Nitrogen-nitrogen double bond.
Explanation:
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