Dysprosium was discovered in 1886. it's freezing point is 1400c and its boiling point is 2600c. if it's specific heat is 0.1733j/g. c, how many joules are required to heat 10.0g from its freezing point to its boiling point?

The correct answer is 12,48 J.

Explanation:

According to the first law of thermodynamics, the variation of the internal energy is equal to the heat minus the work.

ΔU = Q - W

Being:

Q=n * C * (-)

Q: Heat

n: number of moles

C: specific heat

: Final temperature

: Initial temperature

As in this situation there is no variation of the internal energy, the initial equation remains:

W=Q

First, we calculate the number of moles of Dysprosium present in a 10g sample.

n =

n: Number of moles of Dysprosium

m: Mass of Dysprosium

: Molar mass of Dysprosium

n =

n=0,06 mol

Now, replacing the values given in the problem in the equation of Q, we calculate W as:

W = n * C * (-)

W=0,06 mol * 0,1733 * (2600 C - 1400 C)

W=12,48 J

Have a nice day!

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.