Answer:
3.11 mol/kg
Explanation:
Molality M = number of moles of solute, n/mass of solvent, m
To calculate the number of moles of glycerol (C₃H₈O₃) in 22.75 g of glycerol, we find its molar (molecular) mass, M',
So, M' = 3 × atomic mass of carbon + 8 × atomic mass hydrogen + 3 × atomic mass of oxygen
= 3 × 12 g/mol + 8 × 1 g/mol + 3 × 16 g/mol = 36 g/mol + 8 g/mol + 48 g/mol = 92 g/mol.
So, number of moles of glycerol, n = m'/M' where m' = mass of glycerol = 22.75 g and M' = molecular mass of glycerol = 92 g/mol
So, n = m'/M'
n = 22.75 g/92 g/mol
n = 0.247 mol
So, the molality of the solution M = n/m
Since m = mass of ethanol = 79.6 g = 0.0796 kg, substituting the value of n into the equation, we have
M = 0.247 mol/0.0796 kg
M = 3.11 mol/kg
So, the molality of the solution is 3.11 mol/kg.
Answer:
question 3: A
question 4: products and reactants
Explanation:
Answer:
Hello friends
Explanation:
<h3>For a given principal quantum number for or n, the corresponding angular quantum number or is equivalent to a range between 0 and( n-1)</h3>
<h3>This means that the angular quantum number for a principal quantum number of 2 is equivalent to.</h3>
<h3>1 = 0 - > (n - 1) = 0 - > (2 - 1) = 0 - > 1</h3>
<h3>Hope it's helpfully. </h3>
Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.
The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.
When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.
Learn more: brainly.com/question/2510654
In the given examples, Cl2 and CCl4 are non-polar and held by covalent bonds. HI and H2O are polar molecules and held by ionic bonds.
Since covalent bonds are weaker than ionic bonds, Cl2 or CCl4 must have the lowest melting points. Among the two, Cl2 has a lower molecular mass than CCl4, hence Cl2 must have the lowest melting point.
Ans A) Cl2
