Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Answer:
answer is potasium means k
Mn₂O
Explanation:
The oxide that will most likely form colored solutions is Mn₂O.
This is because most transition metals form colored compounds. Manganese is a transition metal belonging to the d-block on the periodic table.
- Other examples of transition metals are scandium, titanium, iron, copper, cobalt, nickel, zinc
- They belong to the d-block on the periodic table.
- They have variable oxidation states.
- Most of their solutions are always colored.
Learn more:
Periodic table brainly.com/question/8543126
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Answer: Option (b) is the correct answer.
Explanation:
In a chemical reaction, the bonds between the reactant molecules tend to break leading to the formation of new bonds to produce products.
So, in order to break the bonds between the reactant molecules, energy is required to overcome the attraction between the atoms.
To form new bonds, energy gets released when two atoms come closer to each other. Hence, formation of bond releases energy.
As in the given reaction it is shown that
< 0, that is, enthalpy change is negative. Hence, energy is released as it is an exothermic process.
Thus, we can conclude that the statement energy released as the bonds in the reactants is broken is greater than the energy absorbed as the bonds in the products are formed, is true about the bond energies in this reaction.
Answer: +178.3 kJ
Explanation:
The chemical equation follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaO%28s%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5E0f_%7BCO_2%7D%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaCO_3%28s%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-635.1%29%29%2B%281%5Ctimes%20%28-393.5%29%29%5D-%5B%281%5Ctimes%20%28-1206.9%29%29%5D)
The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ