Which of these are the primary components in an automobile battery?
1 answer:
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Answer:
Q = 143,921 J = 143.9 kJ.
Explanation:
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In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:
Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:
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Calcium chloride forms an ionic compound with a regularly arranged lattice of oppositely charged ions.
The Na+ and Cl- ions are held together by strong electrostatic forces.
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The Na+ and Cl- ions are held together by strong electrostatic forces.
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Explanation:
Endothermic reactions are chemical reactions in which the reactants absorb heat energy from the surroundings to form products. These reactions lower the temperature of their surrounding area, thereby creating a cooling effect. Physical processes can be endothermic as well – Ice cubes absorb heat energy from their surroundings and melt to form liquid water (no chemical bonds are broken or formed).
When a chemical bond is broken, it is usually accompanied by a release of energy. Similarly, the formation of chemical bonds requires an input of energy. The energy supplied/released can be of various forms (such as heat, light, and electricity). Endothermic reactions generally involve the formation of chemical bonds through the absorption of heat from the surroundings. On the other hand, exothermic reactions involve the release of heat energy generated from bond-breakage.
Endothermic Reaction Examples
Ammonium nitrate (NH4NO3), an important component in instant cold packs, dissociates into the ammonium cation (NH4+) and the nitrate anion (NO3–) when dissolved in water
Answer:
Explanation:
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In this case, the undergoing chemical reaction is:
Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:
Finally, the resulting molarity in 30.8 mL (0.0308 L):
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