Answer:
So X has an atomic mass of 55.05 u, and is probably Mn
Explanation:
<u>Step1: </u>Balance the equation
XI2 +Cl2 → XCl2 + I2
<u>Step2:</u> find the molar mass of the compounds
<em><u>Compounds : </u></em>
X ⇒ Molar mass = M
XI2 ⇒ Molar mass = M + 2*126.9 u
XCl2 ⇒Molar mass = M + 2*35.45u
⇒For 1 mole XI2 we have 1 mole Cl2 reacting , as well as 1 mole XCl2 and 1 mole I2 produced.
<u>Step 3:</u> Calculating the atomic mass
(1.383g * XI2) / (M + 2*126.90 u) = (0.564g * XCl2) / ( M + 2 * 35.45u)
1.383 (M + 2 *35.45 u )= 0.564 (M + 2*126.90)
1.383M + 98.0547 = 0.564 M + 143.1432
1.383 M - 0.564 M =143.1432 - 98.0547
0.819 M = 45.0885
M = 55.05 u
So X has an atomic mass of 55.05 u ⇒ if we look at the periodic table we can find that the closest element is Manganese (Mn)
MnI2 + Cl2 → MnCl2 + I2
They are compressed so that a larger amount of gas can be stored in a smaller container. A greater mass confined to a smaller volume makes transporting and storing of gases easier. Increasing temperature increases pressure, and the cylinders might explode. Before compressed oxygen can be breathed, it must be decompressed.
Explanation:
correct answer is g(x)=x+2
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.