Answer:
Explanation:
When you are in the laboratory and take a direct sniff of the chemicals you are using, you run the risk of damaging your mucous membranes or your lungs. When it is necessary to smell chemicals in the lab, the proper technique is to cup your hand above the container and waft the air toward your face.
Answer:
8.37 grams
Explanation:
The balanced chemical equation is:
C₆H₁₂O₆ ⇒ 2 C₂H₅OH (l) + 2 CO₂ (g)
Now we are asked to calculate the mass of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.
From the ideal gas law we can determine the number of moles that the 2.25 L represent.
From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.
PV = nRT ⇒ n = PV/RT
n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂
Moles glucose required:
0.093 mol CO₂ x ( 1 mol C₆H₁₂O₆ / 2 mol CO₂ ) = 0.046 mol C₆H₁₂O₆
The molar mass of glucose is 180.16 g/mol, then the mass required is
0.046 mol x 180.16 g/mol = 8.37 g
M=7M(H₂O)
M=7*18.015 g/mol = 126.105 g/mol
Because it happens somewhere
2.91 mol Al * ( 26.982 g Al / 1 mol Al) = 78.518 grams
