To make it easier, assume that we have a total of 100 g of a compound. Hence, we have 58.80g of xenon, 7.166g of oxygen, and 34.04g of fluorine.
Know we will convert each of these masses to moles by using the atomic masses:
58.8/131.3 = 0.45 mole of Xe
7.166/16 = 0.45 mole of O
34.04/19 = 1.79 mole of F
Now, we will divide all the mole numbers by the smallest among them and get the number of atoms in the compound:
Xe = 0.45/0.45 = 1
O = 045/0.45 = 1
F = 1.79/0.45 = 3.98 = 4
So, the empirical formula of the compound XeOF₄
Answer:
that is answer
Explanation:
symbol - elements - mass percent
S. - sulfur. - 21.945%
F. - fluorum. - 78.047%
Answer:
[Y₂] = 31.25M
Explanation:
Based on the chemical reaction:
2X(g) + Y2(g) ⇌ 2XY(s)
We must define the equilibrium constant, K as:
K = 20 = 1 / [X]²[Y₂]
<em>Where</em> [X] and [Y₂]<em> are the equilibrium concentrations of each gas.</em>
<em />
If [X] = 4.0x10⁻²M:
20 = 1 / [4.0x10⁻²M]²[Y₂]
0.032 = 1 / [Y₂]
<h3>[Y₂] = 31.25M</h3>
Answer:
Ni
Explanation:
An active metal is a highly reactive metal. Active metals are found high up in the activity series.
Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.
Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.