Answer:
q = -6464.9 kJ
Explanation:
We are given that the heat of combustion is ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.
vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³
m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g
mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol
q = −394 kJ /mol C x 16.41 mol C = -6464.9 kJ
pH is the measure of the hydrogen ion concentration while pOH is of hydroxide ion concentration in the solution. The pH is 0.939 and pOH is 13.061 pOH.
pH is the concentration of the hydrogen ion released or gained by the species in the solution that depicts the acidity and basicity of the solution.
pOH is the concentration of the hydroxide ion in the solution and is dependent on the pH as an increase in pH decreases the pOH and vice versa.
Both HCl and HBr are strong acids and gets ionized 100 % in the solution. If we let 1 L of solution for the acids then the concentration of the hydrogen ion will be 0.100 M.
Since both completely dissociate we would just add the molarities of each of the H+ ions together and then calculate the PH and POH from that :
HCL(0.040M)----> H+(0.040M) +CL-(0.040M)
HBr(0.075M)----> H+(0.075M) +Br-(0.075M)
so 0.040M (H+ from HCL) + 0.075M (H+ from HBr) = 0.115M H+ in total.
pH is calculated as:
pH = -log[H+]
Substituting values in the equation:
log(0.115M)= 0.939 pH
pOH is calculated as:
14 - pH = pOH
Substituting values in the equation above:
14 - 0.939= 13.061 pOH
Therefore, pH is 0.939 and pOH is 13.061.
Learn more about pH and pOH here:
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Well, if you look at group 1 of the periodic table, you will notice a thrend. All elements in group 1 have 1 valence / outer electron. Then you look at period 2, 3, 4 and so on, you will see that the group number corresponds the number of valence/ outershell electrons. Hence, the group determines the electron(s) on the outershell.
No, it won't change the amount of reactants nor the products as a catalyst will only provide an alternative path where lower activation energy is needed for the process to take place.
hope this explains it
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We have Boltzmann's equation S = k ln W
Boltzmann's constant k = 1.381 x 10^-23 J/K
W = Number of absorption sites
At W = 484, Entropy S1 = 1.381 x 10^-23 ln 484 = 8.537 x 10^-23 J/K
At W = 729, Entropy S2 = 1.381 x 10^-23 ln 729 = 9.103 x 10^-23 J/K
Change of Entropy = S2 - S1 = 0.566 x 10^-23 J/K
