First, let's write the chemical formula for each of the substances mentioned in the problem.
Strontium Chloride: SrCl₂
Lithium Phosphate: Li₃PO₄
Strontium Phosphate: Sr₃(PO₄)₂
Lithium Chloride: LiCl
So, the balanced chemical reaction is:
<em>3 SrCl₂ (aq) + 2 Li₃PO₄ (aq) ---> Sr₃(PO₄)₂ (s) + 6 LiCl (aq)</em>
10.861% / 100 = 0.10861
12.428 % / 100 = 0.12428
(0.10861 * 187.9122) + (0.12428 * 190.9407) <span>+ (0.76711 x 192.8938)</span>
= 192.1100 amu .
hope this helps!
Answer:
18.8 g
Explanation:
The equation of the reaction is;
AgClO3(aq) + LiBr(aq)------>LiClO3(aq) + AgBr(s)
Number of moles of AgClO3 = 117.63 g/191.32 g/mol = 0.6 moles
Number of moles of LiBr = 10.23 g/86.845 g/mol = 0.1 moles
Since the molar ratio is 1:1, LiBr is the limiting reactant
Molar mass of solid AgBr = 187.77 g/mol
Mass of precipitate formed = 0.1 moles * 187.77 g/mol
Mass of precipitate formed = 18.8 g
Answer:
A precipitate will be produced
Explanation:
The Ksp of AgBr is:
AgBr(s) → Ag⁺ + Br⁻
5.0x10⁻¹³ = [Ag⁺] Br⁻]
<em>Where [] are the concentrations in equilibrium of each ion.</em>
<em />
And if Q is:
Q = [Ag⁺] Br⁻]
<em>Where the concentrations are actual concentrations of each ion</em>
<em />
We can say:
IF Q >= Ksp, a precipitate will be produced
IF Q < Ksp, no precipitate will be produced.
the molar concentrations are:
[AgNO₃] = [Ag⁺] = 0.002M * (50mL / 100mL) = 0.001M
<em>Because 50mL is the volume of the AgNO₃ solution and 100mL the volume of the mixture of both solutions.</em>
[NaBr] = [Br⁻] = 0.002M * (50mL / 100mL) = 0.001M
Q = [0.001M] * [0.001M]
Q = 1x10⁻⁶
As Q > Ksp,
<h3>A precipitate will be produced</h3>
Answer:
The average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its natural abundance (the decimal associated with percent of atoms of that element that are of a given isotopе). An element does not have an absolute atomic mass.
<em>Hope</em><em> this</em><em> </em><em>helps</em><em> </em><em>:</em><em>)</em>
