The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.
A(t) = A(o) (1/2)^(t/d)
where t is the certain period of time. Substituting the known values,
A(t) = (20 mg)(1/2)^(85.80/14.30)
Solving,
A(t) = 0.3125 mg
Hence, the answer is 0.3125 mg.
Answer:
C₁₂H₂₂O₁₁ + H₂O → C₅H₁₂O₆ + C₆H₁₂O₆
Explanation:
Chemical equation:
C₁₂H₂₂O₁₁ + H₂O → C₅H₁₂O₆ + C₆H₁₂O₆
Source of sucrose:
Sucrose is present in roots of plants and also in fruits. It is storage form of energy. Some insects and bacteria use sucrose as main food. Best example is honeybee which collect sucrose and convert it into honey.
Monomers of sucrose and hydrolysis:
Sucrose consist of monomers glucose and fructose which are join together through glycosidic bond. Hydrolysis break the sucrose molecule into glucose and fructose. In hydrolysis glycosidic bond is break which convert the sucrose into glucose and fructose. Hydrolysis is slow process but this reaction is catalyze by enzyme. The enzyme invertase catalyze this reaction.
The given reaction also completely follow the law of conservation of mass. There are equal number of atoms of elements on both side of chemical equation thus mass remain conserved.
Answer:
If matter is heated and thus its temperature rises more and more, it can be seen that the particles contained in it move ever faster – be it the relatively free movement of the particles in gases or the oscillation around a rest position in solids. The temperature of a substance can therefore be regarded as a measure of the velocity of the particles it contains. With a higher temperature and thus higher particle
Explanation:
The percentage of Chromium in Chromium Oxide is calculated as follow,
Step 1: Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
2(51.99) + 3(16) = 103.98 + 48 = 151.98 u
Step 2: Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
=
× 100
=
68.41 %
So, the %age composition of chromium in chromium oxide is
68.41 %.
K:
m=155g
M=39g/mol
n = 155g / 39g/mol ≈ 3,97mol
KNO₃:
m=122g
M=101g/mol
n = 122g/101g/mol = 1,21mol
2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent
KNO₃ is limiting reagent
