Answer:
Balancing Strategies: To balance this reaction it is best to get the Oxygen atoms on the reactant side of the equation to an even number. Once this is done everything else falls into place. Put a "2" in front of the NaClO3. Change the coefficient in front of the O2.
<u>Answer:</u> The
of the reaction at given temperature is -12.964 kJ/mol.
<u>Explanation:</u>
For the given chemical reaction:

The expression of
for the given reaction:

We are given:

Putting values in above equation, we get:

To calculate the Gibbs free energy of the reaction, we use the equation:

where,
= Gibbs' free energy of the reaction = ?
= Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)
R = Gas constant = 
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
= equilibrium constant in terms of partial pressure = 
Putting values in above equation, we get:

Hence, the
of the reaction at given temperature is -12.964 kJ/mol.
Answer:
0.97 grams
Explanation:
number of molecules=no of moles × Avogadros number
so
number of moles=nu.of molecules ÷ 6.02×10power23 = 1.98×10power21 ÷ 6.02×10p23=0.0033 mol
mass=nu.of moles× molar mass
molar mass of aspartame=
(14×12)+18+(2× 14.0067n)+(5×16)=294.0134 u
mass = 0.0033 ×294.0134 =0.97 gram
3! You have to ensure balance of all the different elements.
Cat urine. May sound weird, may get it wrong, but it's true. Your teacher'll freak out.