Answer:
0.774g of ethanol
0.970mL of ethanol
Explanation:
Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.
In the problem, you need to prepare a 1.2m solution of ethanol (Solute) in t-butanol (solvent).
14.0g of butanol are <em>0.014kg </em>and as you want to prepare the 1.2m solution, you need to add:
0.014kg × (1.2moles / kg) = 0.0168 moles of solute = Moles of ethanol
To convert moles of ethanol to mass you require molar mass (Molar mass ethanol, C₂H₅OH = 46.07g/mol). Thus, mass of 0.0168 moles are:
0.0168moles Ethanol ₓ (46.07g / mol) =
<h3>0.774g of ethanol</h3>
And to convert mass in g to mL you require density of the substance (Density of ethanol = 0.798g/mL):
0.774g ₓ (1mL / 0.798g) =
<h3>0.970mL of ehtanol</h3>
Ans is B.Bromine. Only two elements are liquid at room temperature ( mercury and Bromine) . Helium and chlorine, arsenic are gases.
(Starch) I hope this helps you please check again i'am not sure of my answer
