<span>In a titration, the substance that is unknown and being identified is called analyte. A titration is where a known solution or concentration called the titrant is used to identify and measure an unknown substance which is the analyte.</span>
S + O2 → SO2
<span>z / (32.0655 g S/mol) x (1 mol SO2 / 1 mol S) x (64.0638 g SO2/mol) = (1.9979 z) g SO2 </span>
<span>C + O2 → CO2 </span>
<span>(9.0-z) / (12.01078 g C/mol) x (1 mol CO2 / 1 mol C) x (44.00964 g CO2/mol) = (32.9776 - 3.66418 z) g CO2 </span>
<span>Add the two masses of SO2 and CO2 and set them equal to the amount given in the problem: </span>
<span>(1.9979 z) + (32.9776 - 3.66418 z) = 27.9 </span>
<span>Solve for z algebraically: </span>
<span>z = 3.0 g S</span>
Answer:
the answer is D. Because the 1 atm pressure of water is 40.65 or 40.7.
Answer:
127.3° C, (This is not a choice)
Explanation:
This is about the colligative property of boiling point.
ΔT = Kb . m . i
Where:
ΔT = T° boling of solution - T° boiling of pure solvent
Kb = Boiling constant
m = molal (mol/kg)
i = Van't Hoff factor (number of particles dissolved in solution)
Water is not a ionic compound, but we assume that i = 2
H₂O → H⁺ + OH⁻
T° boling of solution - 118.1°C = 0.52°C . m . 2
Mass of solvent = Solvent volume / Solvent density
Mass of solvent = 500 mL / 1.049g/mL → 476.6 g
Mol of water are mass / molar mass
76 g / 18g/m = 4.22 moles
These moles are in 476.6 g
Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m
T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C
Answer:
C. Lithium
Explanation:
I Goo gled it and I think that's right.
