Answer:
58.9mL
Explanation:
Given parameters:
Initial volume = 34.3mL = 0.0343dm³
Initial concentration = 1.72mM = 1.72 x 10⁻³moldm⁻³
Final concentration = 1.00mM = 1 x 10⁻³ moldm⁻³
Unknown:
Final volume =?
Solution:
Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.
Therefore;
C₁V₁ = C₂V₂
where C and V are concentration and 1 and 2 are initial and final states.
now input the variables;
1.72 x 10⁻³ x 0.0343 = 1 x 10⁻³ x V₂
V₂ = 0.0589dm³ = 58.9mL
The answers would be:
In a solution, the solvent is present in a greater amount.
In a solutions, the solute dissolves in a solvent.
In general, these are the best answers. The solute is what is being dissolved and the solvent is what dissolves. A solvent comes in greater amounts in a solution and it is the dissolving agent.
For example, sugar and water.
To make a sugar water solution, you will need to dissolve sugar in water. Sugar is the solute in this case because it is what is being dissolved. The water is the solvent, because it dissolves the sugar.
If you had more sugar than water, then you cannot make a solution.
Bleach and anytype of cleaner like window cleaner or disinfectant is very dangerous because if you mix those two together it might cause a very deadly fume and can cause lung failure and death within about 5 hours
Answer:
An elementary particle can be one of two groups: a fermion or a boson. Fermions are the building blocks of matter and have mass, while bosons behave as force carriers for fermion interactions and some of them have no mass. The Standard Model is the most accepted way to explain how particles behave, and the forces that affect them. According to this model, the elementary particles are further grouped into quarks, leptons, and gauge bosons, with the Higgs boson having a special status as a non-gauge boson.
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
