In order to determine the concentration of ammonium ions in
the solution prepared by mixing solutions of ammonium sulfate, (NH4)2SO4, and ammonium
nitrate, first calculate the amount of ammonium ions for each solution.<span>
<span>For ammonium sulfate sol'n: 0.360 L x 0.250 mol(NH4)2SO4/ L x 2 mol NH4+ /1 mol(NH4)2SO4 =
0.18 mol NH4+
<span>For ammonium nitrate sol'n: 0.675 x 1.2 mol NH4NO3/L x 1 mol NH4+ /1 molNH4NO3
= 0.81 mol NH4+
Thus, the amount of NH4+ ions is (0.18 + 0.81) mol or 0.99
mol NH4+. To get the concentration, multiply this to the volume of solution
which is assumed to be additive, such that:</span></span></span>
M NH4+ in sol’n = 0.99 mol NH4+/1.035 L = 0.9565 mol NH4+/ L
sol’n
Mole is base unit of substance. The number of particles in one mole is equal to 6.022 × 10²³. The number of particle in a mole is called as Avogadro's number.
The mole of substance is the amount of substance that contain elementary particles as the number of atoms. the symbol of mole is mol. it is the SI unit of amount of substance. The number particles are ion one mole of substances is 6.022 × 10²³. The number of particles in one mole of substance is called as Avogadro's number.
Thus, Mole is base unit of substance. The number of particles in one mole is equal to 6.022 × 10²³. The number of particle in a mole is called as Avogadro's number.
To learn more about moles here
brainly.com/question/26416088
#SPJ1
Answer:
1.31x10⁻³ moles of H₂
Explanation:
This is the equation:
Mg(s) + 2H₂O (g) → Mg(OH)₂ (aq) + H₂(g)
Ratio is 1:1, so 1 mol of Mg is needed to produce 1 mol of H₂
Mass / Molar mass = Mol
0.032 g / 24.3 g/m = 1.31x10⁻³ moles
1.31x10⁻³ moles of H₂(g)
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
