The complete balanced chemical reaction is:
2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S
First let us calculate the number of moles of AgNO3.
moles AgNO3 = 0.315 M * 0.035 L
moles AgNO3 = 0.011025 mol
From the reaction, 1 mole of Na2S is needed for every 2
moles of AgNO3 hence:
moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2
mol AgNO3)
moles Na2S required = 5.5125 x 10^-3 mol
Therefore volume required is:
volume Na2S = 5.5125 x 10^-3 mol / 0.260 M
<span>volume Na2S = 0.0212 L = 21.2 mL</span>
Answer: Temperature
Explanation: Temperature is a measure of average kinetic energy of particles in an object. The hotter the substance, higher is the average kinetic energy of its constituent particles. When we heat a substance, the particles that constitute the substance gain some energy and begin to move faster.
First. let's write the reaction formula: HBr +LiOH ----> LiBr + H₂O
let's get the moles of LiOH first
moles= Molarity x Liters
moles= 0.253 M x 0.01673 Liter= 0.00423 moles LiOH
using the balanced equation, you can see that 1 mol LiOH is equal to 1 mol HBr. so:
0.00423 mol LiOH = 0.00423 mol HBr
now let's find the concentration
molarity= mol/ Liters
0.00423 mol/ 0.01000 Liters= 0.423 M
Answer:
0.912 mL
Explanation:
3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
FeCl3 is the limiting reactant.
Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles
Hence actual yield of Iron III sulphide = 0.043 moles
Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield
Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide
From the reaction equation,
2moles of iron III chloride produced 1 mole of iron III sulphide
x moles of iron III chloride, will produce 0.057 of iron III sulphide
x= 2× 0.057= 0.114 moles of iron III chloride
But
Volume= number of moles/ concentration
Volume= 0.114/0.125
Volume= 0.912 mL
