<h2>The required option d) "specific heat" is correct.</h2>
Explanation:
- To raise the temperature of any substance or material of certain mass to respective temperature it requires some amount of heat.
- Specific heat is the amount of heat necessary to raise the temperature of the substance of 1 gram to 1 Kelvin.
- It is the amount of heat which is required to raise the temperature per unit mass to per unit temperature.
- Thus, the required "option d) specific heat" is correct.
Answer:
1. 0.02 M
2. 0.01 M
3. 4×10⁻⁶
Explanation:
We know that V₁S₁ = V₂S₂
1.
Concentration of HCl = 0.05 M
end point comes at = 10 ml
So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M
2.
2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)
[Ca²⁺] = 0.02 ÷ 2 = 0.01 M
3.
= [Ca²⁺(aq)] [OH⁻(aq)]²
Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)
= [0.01 × (0.02)²] = 4×10⁻⁶
4.
If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.
5.
value describes the solubility of a particular ionic compound. The higher the value, the higher the Solubility will be.
6.
This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility
Answer:
carbon
Explanation:
because it is an allotrope of carbon
Answer:
As for your question, I know to forget to put the options, specifically that your question is incomplete.
Explanation:
Although it could help you by telling you that always a reaction that seeks to balance the pH, and achieve neutrality ... It is necessary to achieve a concentration of OH equal to that of H +, in this way the hydroxyl and the protons.
Answer:
See explanation
Explanation:
In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.
As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.
However, some of the alpha particles were deflected through large angles, in Rutherford's interpretation, the deflected alpha particles had hit the dense positive core of the atom which he called the nucleus.
This accounted for their scattering through large angles throughout the foil in all directions.
