Answer:
a) 6 mol H2O
b) this reaction is endothermic
c) when 1 mol of CO2 is used, in the reaction they occur 0.5025 KJ
Explanation:
balanced eq:
- 6CO2 + 6H2O + 2678 KJ ↔ C6H12O6 + 6O2
6 - C - 6
18 - O - 18
12 - H - 12
a) mol H2O = 6 mol.......from balanced equation.
b) ΔE = 2678 KJ....... this reaction absorbs heat ( ΔE is positive )
c) 1 gramo C6H12O6 ≅ 4 cal
- Mw C6H12O6 = 180.156 g/mol
⇒ 1mol CO2 * ( mol C6H12O6 / 6mol CO2 ) =0.166 mol C6H12O6
⇒ 0.166mol C6H12O6 * ( 180.156 g C6H12O6 / mol ) = 30.026g C6H12O6
⇒30.026 gC6H12O6 * ( 4 cal / gC6H12O6 ) * ( Kcal / 1000 cal ) * (4184 J / Kcal ) * ( KJ / 1000 J ) = 0.5025 KJ C6H12O6.
The given elements put into an equation using their symbols are as follows:
Pb +

=

+ Ag
Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb +

=

+ Ag
Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb +

=

+ Ag
Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb +

=

+ 2Ag
That is your final equation
The coefficients are 2 + 2 = 1 + 2
Answer:
- Last choice: <em><u>- 3.72°C</u></em>
Explanation:
The freezing point depression in a solvent is a colligative property: it depends on the number of solute particles.
The equation to predict the freezing point depression in a solvent is:
Where,
- ΔTf is the freezing point depression of the solvent,
- Kf is the cryoscopic molal constant of the solvent, and i is the Van'f Hoff factor, which is the number of ions produced by each unit formula of the ionic compound.
The calcualtions are in the attached pdf file. Please, open it by clicking on the image of the file.
Answer:
there are two significant figures is the number 8400
Explanation:
I think you want to ask about Keq. At equilibrium, we can know [SO2Cl2] is 2.2*10-2 M -1.3*10-2M=9*10^-3 M. And [SO2]=[Cl2]. So the Keq=1.88*10^-2.