Excess aqueous NaI was added to 50 mL of
1 answer:
The molarity of the solution is 1.69M.
<h3><u>Explanation:</u></h3>The reaction between the two reactants are:
CuNO₃+ NaI = CuI +NaNO₃.
So, 1 mole of CuNO₃ or copper nitrate produces 1 mole of CuI or Copper iodide.
Molecular weight of copper iodide = = 190.4.
So, moles of copper iodide produced here is == 0.08.
So, moles of copper nitrate that was present in the solution = 0.08.
Amount of solution used = 50ml.
So, in 50ml or solution, moles of copper nitrate present = 0.08 moles.
So, in 1 liter of solution, moles of copper nitrate present = 1.69 moles.
So, the molarity of the solution is 1.69M.
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Answer:
Mass of SO₂ can be made from 25.0 g of Na₂SO₃ and 22 g of HCl = 12.672 g
Explanation:
SO₂( sulfur dioxide) can be produced in the lab. by the reaction of hydrochloric acid & sulphite salt such as sodium.
the balanced chemical equation is as follows
Na₂SO₃ + 2 HCl → 2 NaCl + SO₂ + H₂O
Moles of Na₂SO₃ =
Moles of HCl =
using mole ratio method to find limiting reagent
For sodium sulfite
for HCl
since <u>sodium sulfite</u> is <u>limiting reactant</u> for above chemical reaction
1 mole of Na₂SO₃ produce 1 mole of SO₂
0.198 mole of Na₂SO₃ produce 0.198 mole of SO₂
∴ Mass of SO₂ produce = mole x molar mass of SO₂
= 0.198 x 64
= 12.672 g
Answer:
option D is correct
D. This solution is a good buffer.
Explanation:
TRIS (HOCH)
CNH
if TRIS is react with HCL it will form salt
(HOCH)
CNH
+ HCL ⇆ (HOCH
)
NH
CL
Let the reference volume is 100
Mole of TRIS is = 100 × 0.2 = 20
Mole of HCL is = 100 × 0.1 = 10
In the reaction all of the HCL will Consumed,10 moles of the salt will form
and 10 mole of TRIS will left
hence , Final product will be salt +TRIS(9 base)
H = Pk + log (base/ acid)
8.3 + log(10/10)
8.3