Answer:
The answer is in the explanation.
Explanation:
A solution is defined as the <em>homogeneous mixture </em>of a solute (In this case, NaCl) and the solvent (water).
To prepare 1L of the solution, the student can weigh the 3g of NaCl in the volumetric flask but need to add slowly water to dissolve the NaCl (That is very soluble in water). When all NaCl is dissolved the student must transfer the solution to the 1L volumetric flask. Then, you must add more water to the beaker until "Clean" all the solute of the beaker to transfer it completely to the volumetric flask.
The smallest particle of an element that still retains the chemical properties of it is an atom.
The chemical reaction would be expressed as follows:
HBr + LiOH = LiBr + H2O
We are given the volumes and corresponding concentration to be used for the reaction. We use these values to solve for the concentration of the other reactant. We do as follows:
0.253 mol LiOH / L solution ( 0.01673 L ) ( 1 mol HBr / 1 mol LiOH ) = 0.00423 HBr needed
Concentration of HBr =0.00423mol / .010 L = 0.423 M HBr
Answer:
the answer is 30% if the atoms will change to a stable
<span> First you need to know how many isotopes there are of silicon, and its average atomic units (look at periodic table). Then make up a system of equations to solve for it. Theres 3 stable silicon isotopes (28, 29, 30) so you will need to have 3 equations. You must be given the percent abundance of at least one of the isotopes to solve because here I can only see 2 equations (numbered down below) set x = percent abundance of si-28 y = percent abundance of si-29 z = percent abundance of si-30 since all of silicon atoms account for 100% of all silicon: x + y + z = 100% = 1 therefore: 1) x = 1 - y - z You also have 2) 28x + 29y + 30z = average atomic mass you can substitute x so that equation becomes: 28 (1 - y - z) + 29y + 30z = average atomic mass See how you have 2 variables here? You cant go on until you know the value of one isotope already or you have given a clue which you can derive the third equation</span>
