Answer:-
0.91% is the students % of error
Explanation: -
Accepted value= 12.11 grams
Measured value = 12.22 grams
Error = 12.22-12.11 = 0.11 grams
Percentage error =
x100
= 0.91 %
Thus 0.91% is the students % of error
Given mass of tungsten, W = 415 g
Molar mass of tungsten, W = 183.85 g/mol
Calculating moles of tungsten from mass and molar mass:

Answer is: the boiling point of the resulting solution of sucrose is 100.42°C.
m(H₂<span>O) = 15.2 g ÷ 1000 g/kg = 0.0152 kg.
</span>m(C₁₂H₂₂O₁₁<span>) = 4.27 g.
n</span>(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 4.27 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0125 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0125 mol ÷ 0.0152 kg.
b(solution) = 0.82 m.
ΔT = b(solution) · Kb(H₂O).
ΔT = 0.82 m · 0.512°C/m.
ΔT = 0.42°C.
Tb = 100°C + 0.42°C = 100.42°C.
Instrumental methods of analysis rely on machines.The visualization of single molecules, single biological cells, biological tissues and nanomaterials is very important and attractive approach in analytical science.
There are several different types of instrumental analysis. Some are suitable for detecting and identifying elements, while others are better suited to compounds. In general, instrumental methods of analysis are:
-Fast
-Accurate (they reliably identify elements and compounds)
-Sensitive (they can detect very small amounts of a substance in a small amount of sample)