<span>Dmitri Mendeleev's Periodic Law states that chemical and physical properties repeat themselves in groups of 8.</span>
From the Graham's law of effusion;
R1/R2 = √MM2/√MM1
Molar mass of chlorine gas is 71
Therefore;
1.87= √ 71 /√mm1
= 1.87² = 71/mm1
mm1 = 71/1.87²
= 71/3.4969
= 20.3
Thus, the molar mass of the other gas is 20.3 , and i think the gas is neon
Explanation:
The balanced equation of the reaction is given as;
Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)
1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?
From the reaction;
2 mol of HBr produces 1 mol of MgBr2
Converting to masses using;
Mass = Number of moles * Molar mass
Molar mass of HBr = 80.91 g/mol
Molar mass of MgBr2 = 184.113 g/mol
This means;
(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2
18.3g would produce x
161.82 = 184.113
18.3 = x
x = (184.113 * 18.3 ) / 161.82 = 20.8 g
2. How many moles of H2O will be produced from 18.3 grams of HBr?
Converting the mass to mol;
Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol
From the reaction;
2 mol of HBr produces 2 mol of H2O
0.226 mol would produce x
2 =2
0.226 = x
x = 0.226 * 2 / 2 = 0.226 mol
3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?
From the reaction;
2 mol of HBr reacts with 1 mol of Mg(OH)2
18.3g of HBr = 0.226 mol
2 = 1
0.226 = x
x = 0.226 * 1 /2
x = 0.113 mol
Answer:

Explanation:
Formula for the calculation of no. of Mol is as follows:

Molecular mass of Ag = 107.87 g/mol
Amount of Ag = 5.723 g

Molecular mass of S = 32 g/mol
Amount of S = 0.852 g

Molecular mass of O = 16 g/mol
Amount of O = 1.695 g

In order to get integer value, divide mol by smallest no.
Therefore, divide by 0.02657



Therefore, empirical formula of the compound = 
Answer:
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