Hrxn = Q reaction / mol of reaction
mol of reaction = M * V = 10 * 1 = 10 mmol = 0.01 mol
Q water = m * C * (Tf - Ti)
= (10 + 10) (4.184) (26-20) = 502.08 J
Q reaction = - Q water = -502.08 J
Hrxn = -502.08 / (0.01) = - 50208 J = - 50.21 kJ/mol
Answer:
A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)
Explanation:
Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.
The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:
Ca(s), C(s) and O₂(g)
That means, the equation that represents standard enthalpy of CaCO₃ is:
<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />
<em>Is the equation that has ΔH° = -1207kJ/mol</em>
Depends on what the base is. You would reference the base dissociation chart for that value.
Answer:
T<span>he gaseous product of this reaction is water (Option-A).
Explanation:
This is a very interesting experiment. Take sugar in a beaker and add concentrated Sulfuric Acid into it. After a while an exothermic reaction will initiate with the formation of Carbon Black and Water vapors. You will observe the formation of hard and hot stem like body which is completely Black. This blackness is due to C and the water vapors will eliminate in the form of steam as the temperature has arised.</span>
