Answer:
ΔH°rxn = -47 kJ
Explanation:
Using Hess´s law for the reaction:
3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ,
the ΔH°rxn will be given by the expression:
ΔH°rxn kJ = 2ΔHºf(Fe3O4) + ΔHºf(CO2) - ( 3ΔHºf(Fe2O3) + ΔHºf(CO) )
= 2(-1118) + (-394) - ( 3( -824 ) + ( -111 ) )
= - 47 kJ
Answer: 4 : 3
Explanation:Please see attachment for explanation
The question here is solved using basic chemistry. CaCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of CaCl2 that dissolves.
CaCl2(s) --> Ca+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.65 mol CaCl2/1L × 2 mol Cl⁻ / 1 mol CaCl2 = 1.3 M
The answer to this question is [Cl⁻] = 1.3 M
Answer:
Fe
Explanation:
The cell potential is:
ΔE°cell = E°red(red) - E°red(oxid)
Where, E°red(red) is the reduction potential of the substance that is reducing, and E°red(oxid) is the reduction potential of the substance that is oxidizing. For the reaction be spontaneous and happen, ΔE°cell > 0.
The reduction takes place in the cathode, which is the negative pole, and the oxidation in the anode, which is the positive pole. So, the electrons flow from the positive pole to the negative pole (anode to cathode).
Then, if the voltmeter measured a negative potential, it means that is was attached incorrectly. So, the anode is Fe.
You can use the follwing equations.
pH=14-pOH
pOH=-log[OH⁻]
a) pOH=-log10⁻¹²
pOH=12
pH=14-12
pH=2
You can also use these equations:
[H⁺]=K(w)/[OH⁻] (K(w)=1.01×10⁻¹⁴)
pH=-log[H⁺]
b) [H⁺]=(1.01×10⁻¹⁴)/10⁻²M
[H⁺]=1.01×10⁻¹²M²
pH=-log(1.01×10⁻¹²)
pH=12
You can use either method. It does not really matter.
I hope this helps. Let me know if anything is unclear and when you do the calculations for c you should get pH=7.