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
Explanation:
clinical laboratory test results are a very important parameter in diagnosis, monitoring and screening. 70-80 ... it is possible to determine whether you can interpret the test result as being ...
1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution
=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution
2) Molarity = M = number of moles of solute / liters of solution
=> calculate the number of moles of 8.9 grams of NaCl
3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol
4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol
5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M
Answer: 0.152 M
Answer:
It is heterogeneous as it is made up of visibly different substances or phases.
Answer:
333.3mL
Explanation:
Using the formula as follows:
C1V1 = C2V2
Where;
C1 = initial concentration (M)
C2 = final concentration (M)
V1 = initial volume (mL)
V2 = final volume (mL)
According to the information provided in this question,
C1 = 4.00M
C2 = 1.50M
V1 = 125mL
V2 = ?
Using C1V1 = C2V2
4 × 125 = 1.5 × V2
500 = 1.5V2
V2 = 500/1.5
V2 = 333.3mL
Therefore, the CuSO4 solution needs to be diluted to 333.3mL to make 1.50 M solution.