The initial sample has a molecular formula of MnSO₄·H₂O. This molecule is a hydrate as it has a unit of water within its structure for every molecule of MnSO₄. This sample is being dehydrated to remove the water to give.
MnSO₄·H₂O → MnSO₄ + H₂O
MnSO₄·H₂O has a molecular mass of 169.02 g/mol. While MnSO₄ has a molecular mass of 151 g/mol. Water has a molecular mass of 18.02 g/mol. We now can use the ratio of the mass of water to the mass of the initial sample to determine the percentage of each component by mass.
% water by mass:
18.02/169.02 x 100% = 10.7% Water by mass.
% MnO₄ by mass:
151/169.02 x 100% = 89.3% MnSO₄ by mass.
Water makes up 10.7% of the initial mass of MnSO₄·H₂O.
Answer:
The answer to the question is
The height of the mercury fluid column remain the same.
Explanation:
The pressure, P in a column of fluid of height, h is given by
P = (Density p)×(height of fluid column h)×(gravity g)
Therefore, when the diameter is doubled we have
Density of the mercury in the tube with twice the diameter = (Mass of mercury)/(volume of mercury) where the volume of mercury = h×pi×(Diameter×2)^2/4 = h×pi×Diameter^2. Therefore the volume increases by a factor of 4 and therefore the mass increases by a factor of 4 which means that the density remains the same hence
P = p×h1×g = p×h2×g Therefore h1 = h2
The height of the fluid column remain the same
The answer is oxidation.
That is in the redox fueling reaction,
succinate + NAD ↔fumarate + NADPH, the succinate molecule is undergoing oxidation.
As succinate molecule is providing electrons to NAD, so that it can be reduced from NAD to NADPH. So it is losing electrons and undergoing oxidation.
So the answer is oxidation.
1 is an element, 2 is an compound
