Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
Answer:
The answer to your question is V2 = 434.7 l
Explanation:
Data
Volume 1 = V1 = 240 l Volume 2 = ?
Temperature 1 = T1 = 479°K Temperature 2 = T2 = 293°K
Pressure 1 = P1 = 300 KPa Pressure 2 = P2 = 101.325 Kpa
Process
1.- Use the combined gas law to solve this problem
P1V1/T1 = P2V2/t2
-Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (300)(240)(293) / (479)(101.325)
3.- Simplification
V2 = 21096000 / 48534.675
4.- Result
V2 = 434.7 l
Sodium carbonate contains 3 elements: Na, C, and O. Step 2: Determine how many atoms of each element are in the formula. Na, Co, is made up of Na, C, and O3, so Na.co, has 2 atoms of sodium (Na) 1 atom of carbon (C) and 3 atoms of oxygen (O).
Electrical?
Hope this helps