Heat gained by ice cubes would be equal to the - heat lost by warm water
The moles of ice is: 50.5 g / 18.0 g/mol = 2.81 mol
Heat required to melt all of the ice is equal to: 2.81 mol X 6.02 kJ/mol = 16.9 kJ = 16890 J
Now, know whether the warm water will still be above 0C when it loses this much heat:
-1690 J = 160 g (4.184 J/gC) (Delta T) Delta T = -25C
In order to solve for the final temperature, going back to include warming of the melted ice to a final temperature:
q(ice/water) = - q(warm water)
moles (Delta Hf) + m c (T2-T1) = - m c (T2-T1)
50.5 g / 18.0 g/mol = 2.81 mol
2.81 mol X 6.02 kJ/mol + 50.5g (4.184 J/gC) (T2-0) = -160g (4.184 J/gC) ( T2-80)
16916 + 211.3T2 = -669.4 T2 + 53555
36639 = 880.7 T2
T2 = 41.6 C
Answer:
There is this property called CATENATION which is basically the tendency of the atoms to bind to eachother. It is the MOST in the carbon after that silicon.
So the answer should Carbon atoms bond together to form groups of compounds.
Q is the heat added to the system.
Since it is stated that it is an ideal gas, we use the ideal gas equation to solve the volume of this gas sample. The ideal gas equation is expressed as:
PV = nRT
V = nRT / P
V = 0.200 (8.314) (400) / 200x10^3
V = 3.33 x 10^-3 m³ or 3.33L
Therefore, the correct answer is option B.
Answer:
pH = 1.85
Explanation:
The reaction of H₂NNH₂ with HNO₃ is::
H₂NNH₂ + HNO₃ → H₂NNH₃⁺ + NO₃⁻
Moles of H₂NNH₂ and HNO₃ are:
H₂NNH₂: 0.0400L ₓ (0.200mol / L) = 8.00x10⁻³ moles of H₂NNH₂
HNO₃: 0.1000L ₓ (0.100mol / L) = 0.01 moles of HNO₃
As moles of HNO₃ > moles of H₂NNH₂, all H₂NNH₂ will react producing H₂NNH₃⁺, but you will have an excess of HNO₃ (Strong acid).
Moles of HNO₃ in excess are:
0.01 mol - 8.00x10⁻³ moles = 2.00x10⁻³ moles of HNO₃ = moles of H⁺
Total volume is 100.0mL + 40.0mL = 140.0mL = 0.1400L.
Thus, [H⁺] is:
[H⁺] = 2.00x10⁻³ moles / 0.1400L = 0.0143M
As pH = - log [H⁺]
<h3>pH = 1.85 </h3>
