Initially, we have a number of moles (n₁) a gas sample at a certain pressure (P), temperature (T₁) and volume (V). We can relate these variables through the ideal gas equation.

P . V = n₁ . R . T₁

where,

R is the ideal gas constant

We can rearrange this equation like:

$T_{1}=\frac{P.V}{n_{1}.R}$

If only one fourth of the initial molecules remain n₂ = 1/4 n₁. The new temperature (T₂) assuming pressure and temperature remain constant is:

$P.V=n_{2}.R.T_{2}=\frac{1}{4} n_{1}.R.T_{2}\\\frac{P.V}{n_{1}.R} =\frac{1}{4} T_{2}\\T_{1}=\frac{1}{4} T_{2}\\T_{2}=4.T_{1}$

6 0

3 years ago

Why doesn't potassium fluoride conduct electricity as a solid?

Korvikt [17]

potassium flouride is an ionic compound and ionic compounds do not conduct electricity in solid form instead it conducts electricity in molten state.Ionic compounds do not conduct electricity in solid state because their ions are fixed while in molten form the ions are free

5 0

3 years ago

9) Give the set of four quantum numbers that could represent the extra electron added (using the Aufbau principle) to the neutra

Pie

Answer:

See explanation

Explanation:

9) If an extra electron is added to the neon atom, then the electronic configuration becomes; 1s2 2s2 2p6 3s1

This last electron has quantum numbers;

n=3, l=0, m=0 and s = +1/2

This is so because the 2s level is already filled so the extra electron must go into the 3s level. The orbital quantum number and the magnetic quantum number for the s orbital is zero.

10) Electron affinity is the energy released when one mole of gaseous atoms accept one mole of gaseous electrons to form one mole of gaseous ions having a negative charge.

In the second period, fluorine has the greatest electron affinity since electron affinity increases across the period. The noble gas, neon has an electron affinity of 0KJ/mol.

11) Ionization energy decreases down the group but increases across the period due to increase in the size of the nuclear charge and decrease in the distance between the nucleus and the outermost electron. Hence, the process; Br+(g) ---->Br2+(g) + e- has the greatest ionization energy. Recall that the second ionization is always higher than the first ionization energy.

12)The order of decreasing metallic character here is K> As> P. Even though As and P belong to the same group, we must note that metallic character increases down the group hence the order written above.

3 0

3 years ago