Answer:
A. 0.2395 w/w %
B. 2394ppm
Explanation:
A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:
Mass glycerol / Total mass * 100
<em>Mass glycerol:</em>
The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):
2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol
<em>Mass of water:</em>
998.9mL and density = 0.9982g/mL:
998.9mL * (0.9982g/mL) = 997.1g of water.
That means percent by mass is:
% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %
B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:
2.394g * (1000mg / 1g) = 2394mg:
Parts per million: 2394mg / L = 2394ppm
The answer is c because it is helpful to understand the balance between these two ions. An aqueous solution is a solution in which water is the solvent. Water molecules (H2O) are polar, meaning that they have a negative end (the oxygen) and a positive end (the hydrogens). When there is a reaction in an aqueous solution, the water molecules have the ability to attract and temporarily hold a donated proton (H+). This creates the hydronium ion (H3O+).
=Spectral lines are produced by transitions of electrons within atoms or ions. As the electrons move closer to or farther from the nucleus of an atom (or of an ion), energy in the form of light (or other radiation) is emitted or absorbed.…
Answer:
Number 1. The effective nuclear change of oxygen is greater that of fluorine.
Answer:
8.757 x 10^23 molecules
Explanation:
Use Avogadro's number of 6.022
×
10
^23 molecules per mole.
64g CO2/44.01 g/mol = 1.454 moles of CO2
1.454 x 6.022
×
10
^23 =
8.757 x 10^23 molecules
