Answer:
2.03 moles of Gold
Explanation:
Gold is one of the most precious metal metal used in many applications and mainly as a jewellery. In terms of purity it is categorized in Karats. 24 Karat is considered the purest Gold (i.e. 100 % Gold) while other Karats (14, 18, 22 e.t.c) are alloys with other metals and gyms.
Data Given:
Mass of Gold = 400 g
A.Mass of Gold = 196.97 g.mol⁻¹
Calculate Moles of Gold as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 400 g ÷ 196.97 g.mol⁻¹
Moles = 2.03 moles of Gold
Answer:
The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>
Explanation:
The formula for molal boiling Point elevation is :
= elevation in boiling Point
= Boiling point constant( ebullioscopic constant)
m = molality of the solution
<em>i =</em> Van't Hoff Factor
Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .
In solution Mg3(PO4)2 dissociates as follow :
Total ions after dissociation in solution :
= 3 ions of Mg + 2 ions of phosphate
Total ions = 5
<em>i =</em> Van't Hoff Factor = 5
m = 8.5 m
= 0.512 °C/m
Insert the values and calculate temperature change:
Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K
= 373.15 K[/tex]
21.76 = T - 373.15
T = 373.15 + 21.76
T =394.91 K
Answer:
aldehyde
carbon-1
ketone
carbon-2
Explanation:
Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.
In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.
In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.
In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.
Methylene blue indicates the presence of oxidizing agents because it is oxidized itself by these compounds. When electrons are stripped from methylene blue, the resulting molecule imparts a blue color to the solution--giving a clear sign of a chemical change.