Answer:
146 kJ
Explanation:
There are two heat flows in this question.
Heat lost on cooling + heat lost on solidifying = 0
q₁ + q₂ = 0
mCΔT + nΔHsol = 0
Data:
m = 575 g
C = 0.449 J·K⁻¹g⁻¹
T_i = 1825 K
T_f = 1811 K
ΔHsol = -13.8 kJ·mol⁻¹
Calculations:
(a) Heat lost on cooling
ΔT = T_f - T_i = 1811 K - 1825 K = -14 K
q₁ = mCΔT = 575 g × 0.449 J·K⁻¹g⁻¹ × (-14 K) = -361 J = -3.61 kJ
(b) Heat lost on solidifying
(c) Total heat lost
q = q₁ + q₂ = -3.61 kJ - 142.1 kJ = -146 kJ
The heat lost was 146 kJ.
Elements in the same group have similar properties
Answer:
Increasing the pressure on a reaction involving reacting gases increases the rate of reaction. Changing the pressure on a reaction which involves only solids or liquids has no effect on the rate.
Explanation:
Answer:
Chemoautotrophs
Explanation:
Autotrophs are groups of organisms that are capable of manufacturing their own food (organic molecules) through the fixation of carbon dioxide. There are two types of autotrophs:
- <em>Photoautotrophs fix carbon dioxide by using light as the energy for driving the process.</em>
- <em> </em><em>Chemoautotrophs f</em><em>ix carbon dioxide by using energy from the oxidation of inorganic molecules such as magnesium, or sulfur.</em>
Chemoautotrophs usually inhabit extreme environment such hot vents, deep sea, etc.
The balanced equation for the reaction is as follows
C₆H₁₂O₆ + 6O₃ --> 6CO₂ + 6H₂O
stoichiometry of glucose to CO₂ is 1:6
when 1 mol of glucose reacts - 6 mol of CO₂ are formed
therefore when 5.87 mol of glucose reacts - 6 x 5.87 mol = 35.22 mol
therefore 35.2 mol of CO₂ is formed
