Answer:
See Explanation
Explanation:
For SF6;
Since;
1.25 g of S corresponds to 4.44g of F
1 g of sulphur corresponds to 1 * 4.44/1.25 = 3.55
For SF4;
Since;
1.88 g of S corresponds to 4.44g of F
1 g of sulphur corresponds to 1 * 4.44/ 1.88 = 2.36
Hence;
Mass of oxygen per gram of sulphur in SF6/Mass of oxygen per gram of sulphur in SF4
=
3.55/2.36 = 1.5
Hence the law of multiple proportion is obeyed here.
Answer:
227.78g of the precipitate are produced
Explanation:
Based on the reaction, 3 moles of CuCl2 produce 1 mole of Cu3(PO4)2 (The precipitate).
To solve this question we need to find the moles of CuCl2 added. With these moles and the reactio we can find the moles of Cu3(PO4)2 and its mass as follows:
<em>Moles CuCl2:</em>
285mL = 0.285L * (6.3mol / L) = 1.7955 moles CuCl2
<em>Moles Cu3(PO4)2:</em>
1.7955 moles CuCl2 * (1mol Cu3(PO4)2 / 3mol CuCl2) = 0.5985 moles Cu3(PO4)2
<em>Mass Cu3(PO4)2 -380.58g/mol-</em>
0.5985 moles Cu3(PO4)2 * (380.58g/mol) =
227.78g of the precipitate are produced
Explanation:
I need a help of math.
please can someone help me ?
To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
V2 = P1 V1 / P2
V2 = 153 x 3.00 / 203
<span>V2 = 2.26 L</span>
Amount of oxygen in the compound = 160 g
Amount of oxygen in the compound = 20.2 gm
Mole of oxygen in the compound = 160/16
= 10 moles
Mole of hydrogen in the compound = 20.2/1.01
= 20 moles
Then
The ratio of oxygen to ration of hydrogen = 1:2
So
The empirical formula of the compound is H2O. I hope the answer has come to your help.
