Answer:
Q = 4.056 J
Explanation:
∴ m = 406.0 mg = 0.406 g
∴ <em>C </em>= 1.85 J/g.K
∴ T1 = 33.5°C ≅ 306.5 K
∴ T2 = 38.9°C = 311.9 K
⇒ ΔT = 311.9 - 306.5 = 5.4 K
⇒ Q = (0.406 g)(1.85 J/gK)(5.4 K)
⇒ Q = 4.056 J
Answer: option d. all of the above.
Explanation:
A mineral is an element or a inorganic compound that existes in nature as solid cristals; usually combined with other minerals in ores.
Some examples of minerals, among many, are titania, wich is TiO2, zirconia, which is ZrO2, silica, which is SiO2, gold, Au, silver, Ag.
As you see the definition and examples given meet the whole features included in the stament: a. the have a chemical formula, b they occur naturally, and c.have a characteristic internal structure (that is the way how the atoms are arranged in the specifi cristal).
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then, 

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.