Answer:
The specific heat of iron is 0.45 J/g.°C
Explanation:
The amount of heat absorbed by the metal is given by:
heat = m x Sh x ΔT
From the data, we have:
heat = 180.8 J
mass = m = 22.44 g
ΔT = Final temperature - Initial temperature = 39.0°C - 21.1 °C = 17.9°C
Thus, we calculate the specific heat of iron (Sh) as follows:
Sh = heat/(m x ΔT) = (180.8 J)/(22.44 g x 17.9°C) = 0.45 J/g.°C
It should be 3 feet... I did something like this one day in school but I’m trying to remember
Answer:
The answer is letter A. Diamond
Explanation:
The diamond is a crystalline form of carbon, with a cubic structure, of the sp³ type, that is, each carbon atom of the structure is linked to four other carbon atoms, forming a tetrahedral geometry.
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.
Answer: penetration is the ability of an electron in a given orbital to approach the nucleus closely. Shielding refers to the fact that core electrons reduce the degree of nuclear attraction felt by the orbital electrons. Shielding is the opposite of penetration. The most penetrating orbital is the least screening orbital. The order of increasing shielding effect/decreasing penetration is s<p<d<f.
Explanation:
The order of penetrating power is 1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f....
Since the 3p orbital is more penetrating than the 3d orbital, it will lie nearer to the nucleus and thus possess lower energy.