Answer:
Barium nitrate or silver nitrate based on the anion our solute contains
Explanation:
I assume the situation is that currently the solute is soluble in water and you wish to make it insoluble.
It really depends on the soluble material you have, however, let's look at some specific cases.
- We have a salt in our solution. Addition of any of the three reagents will produce a double displacement reaction, that is, our cation will be replaced by another cation, either sodium, barium or silver cation.
- According to the solubility rules, all sodium salts are soluble, so sodium nitrate won't precipitate our anion.
- In case our solute contains sulfate, carbonate or phosphate, we may use barium nitrate to precipitate it, as barium sulfate, barium carbonate and barium phosphate are insoluble.
- In case our solute contains chloride, then silver nitrate is the way to go to precipitate it in an insoluble form of AgCl. Similarly, silver would form precipitates with carbonate, phosphate, iodide, bromide and slightly soluble silver sulfate (barium is the choice for sulfate, however).
Answer:
When excited electrons fall to lower energy levels, they can release energy in the form of light. metal ions in the salts used in the flame tests.
Answer:
genes (c)
Explanation:
since traits are passed down genetically, so her mother or father must have one (or even one of her grandparents)
hope this helps!:)
1)
<span>m(NaCl) = 1.95 g
V(H2O) = 250mL
M(NaCl) = </span><span>58.5 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
</span>V(H2O) = 250ml = 250g = 0.25 kg<span>
</span><span>molality of NaCl:
</span><span>
n(NaCl)=m/M=1.95/58.5= 0.033 mole
</span>molality b(NaCl)=n(NaCl) / V (H2O)= 0.033/0.25 = 0.132 mol/kg
<span>
milimolality of NaOH = 0.132/0,001 = 132 mmole/kg
</span>
milliosmolality of NaOH = milimolality x N of ions formed in dissociation
Since NaCl dissociates into 2 ions in solution:
<span>
</span>milliosmolality of NaOH = 132 x 2 = 264 osmol<span>es/kg
</span>
2)
m(gl) = 9 g
V(H2O) = 250mL
M(NaCl) = 180 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
V(H2O) = 250ml = 250g = 0.25 kg
molality of glucose:
n(gl)=m/M=9/180= 0.05 mole
molality b(gl)=n(gl) / V (H2O)= 0.05/0.25 = 0.2 mol/kg
milimolality of glucose = 0.132/0,001 = 200 mmole/kg
milliosmolality of glucose = milimolality x N of ions formed in dissociation
Since glucose does not dissociate, milimolality and milliosmolality are same:
milliosmolality of glucose = 200 osmoles/kg
3)
The osmosis represents the diffusion of solvent molecules through a semi-permeable membrane that allows passage solvent molecules but does not to the dissolved substance molecule. The osmosis occurs when the concentrations of the solution on both sides of the membrane are different. Since the semi-permeable membrane only permeates the solvent molecules, but not the particles of the dissolved substance, it occurs the solvent diffusion through the membrane, i.e. the solvent molecules pass through the membrane to equalize the concentration on both sides of the membrane. Solvents molecules move from the middle with a lower concentration in the middle with a higher concentration of dissolved substances.
In our case, osmosis will occur because the concentration of NaCl solution and the concentration of glucose solution do not have same values. Osmosis will occur in the direction of glucose solution because it has a lower concentration.
Find the molar mass of CaCO3 then subtract the molar mass what it originally weighed and the loss of mass. Hopefully this works!
