Answer:
%Sgr = 1% (1 sig.fig.)
Explanation:
mass water = 350ml x 0.975g/ml = 341.25g
mass sugar added = 4g
solution mass = 341.25g + 4g = 345.25g
%sugar = (4g/345.25g)·100% = 1.1586% ≅ 1% (1 sig.fig)
2.5 x 10^24 molecules of C4H10
Answer:
We need 7.5 mL of the 1M stock of NaCl
Explanation:
Data given:
Stock = 1M this means 1 mol/ L
A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL
Step 2: Calculate the volume of stock we need
The moles of solute will be constant
and n = M*V
M1*V1 = M2*V2
⇒ with M1 = the initial molair concentration = 1M
⇒ with V1 = the volume we need of the stock
⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L
⇒ with M2 = the concentration of the new solution = 0.15 M
1*V1 = 0.15*(50)
V1 = 7.5 mL
Since 0.0075 L of 1M solution contains 0.0075 moles
50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M
We need 7.5 mL of the 1M stock of NaCl
Answer:
A
Explanation:
I took a test like this and my teacher said this was right so ye. hopefully I remembered correctly if its wrong im sorry ;C