No. When water first begins to cool down, it contracts. However, as it gets colder and eventually freezes, it begins to expand.
You can test this by freezing water in a water bottle: when you take it out of the freezer, the cap might have popped off or cracks may have formed in the sides of the bottle.
Answer: Water expands when frozen, not contracts.
We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.
150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo
1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo
Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.
Answer:
0.9%
Explanation:
At the instance where you add RBCs to a solution of 0.9% NaCl, they will maintain their shape as well as their size, this is simply to indicate an equilibrium. That entails that the solution is isotonic to the RBCs. This must mean that the concentration of solutes on each side must be equal otherwise osmotic pressure would force water either in or out of the membrane. Therefore, one can also safely assume that RBC's also have a solute concentration of 0.9%.
Answer:
87.9%
Explanation:
Balanced Chemical Equation:
HCl + NaOH = NaCl + H2O
We are Given:
Mass of H2O = 9.17 g
Mass of HCl = 21.1 g
Mass of NaOH = 43.6 g
First, calculate the moles of both HCl and NaOH:
Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles
Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles
Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:
Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles
Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles
From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:
Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g
% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%
