mole=10 x 10⁻³ : 46 g/mol = 2.17 x 10⁻⁴
"if it is tested in a controlled setting with repeated results" is the statement among the choices given in the question that best describes that can possibly make this scientific claim valid. The correct option among all the options that are given in the question is the first option or option "A". I hope the answer has helped you.<span>
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Answer:
0.133 mol (corrected to 3 sig.fig)
Explanation:
Take the atomic mass of H=1.0, and O=16.0,
no. of moles = mass / molar mass
so no. of moles of H2O produced = 1.2 / (1.0x2+16.0)
= 0.0666666 mol
From the equation, the mole ratio of H2:H2O = 2:2 = 1:1,
meaning every 1 mole of H2 reacted gives out 1 mole of water.
So, the no, of moles of H2 required should equal to the no, of moles of H2O produced, which is also 0.0666666 moles.
mass = no. of moles x molar mass
hence,
mass of H2 required = 0.066666666 x (1.0x2)
= 0.133 mol (corrected to 3 sig.fig)
Answer:
Density
Explanation:
Density is defined as the mass per unit volume. It is the ratio between the mass and the volume of a substance. It does not matter how large or small a sample of matter is, the same substance will always have the same density, because of this. The ratio between the mass and volume remains the same.
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
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