Answer:62.66°C or 235.66K
Explanation:Q=McpT, the energy was given in calories so you first convert to Joules by multiplying the value in calories by 4.184J.
17*4.184=71.128kJ.
71.128kJ=mcpT
71.128kJ=245*4.187*(T-Tm)
Tm is the final temperature of the mixture. The T is the temperature given which should be converted to Kelvin by adding 273...T=32+273=305K.
71128J=245*4.187*(305-Tm)
71128=312873.575-1025.815Tm
1025.815Tm=312873.575-71128
1025.815Tm=241745.58
Tm=241745.58/1025.815
Tm=235.66K
Answer:
Kc for this reaction is 0.06825
Explanation:
Step 1: Data given
Number of moles formaldehyde CH2O = 0.055 moles
Volume = 500 mL = 0.500 L
At equilibrium, the CH2O(g) concentration = 0.051 mol
Step 2: The balanced equation
CH2O <=> H2 + CO
Step 3: Calculate the initial concentrations
Concentration = moles / volume
[CH2O] = 0.055 moles . 0.500 L
[CH2O] = 0.11 M
[H2] = 0M
[CO] = 0M
Step 4: The concentration at the equilibrium
[CH2O] = 0.11 - X M = 0.051 M
[H2] = XM
[CO] = XM
[CH2O] = 0.11 - X M = 0.051 M
X = 0.11 - 0.051 = 0.059
[H2] = XM = 0.059 M
[CO] = XM = 0.059 M
Step 5: Calculate Kc
Kc = [H2][CO]/[CHO]
Kc = (0.059 * 0.059) / 0.051
Kc = 0.06825
Kc for this reaction is 0.06825
pH=6.98
Explanation:
This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.
As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.
2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!
At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have
KW=[H3O+]⋅[OH−]=10−14
Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have
[H3O+]=√10−14=10−7M
The pH of pure water will thus be
pH=−log([H3O+])
pH=−log(10−7)=7
Now, let's assume that you're working with a 1.0-L solution of pure water and you add some 10
