The formula for mole fraction is:
-(1)
The solubility of oxygen gas = 1.0 mmol/L (given)
1.0 mmol/L means 1.0 mmol are present in 1 L.
Converting mmol to mol:
So, moles of oxygen = 0.001 mol
For moles of water:
1 L of water = 1000 mL of water
Since, the density of water is 1.0 g/mL.
So, the mass of water is 1000 g.
Molar mass of water = 18 g/mol.
Number of moles of water =
Substituting the values in formula (1):
Hence, the mole fraction is .
Since intermolecular forces is stronger in a solid than a liquid, then NaCl has stronger intermolecular forces than H2O.
Intermolecular forces exists between molecules of the same or different kinds. Water has a dipole moment hence it can interact effectively with the ions in NaCl. This leads to an ion dipole interaction that results in the dissolution of NaCl in H2O.
NaCl is a solid and water is a liquid. Since intermolecular forces is stronger in a solid than a liquid, then NaCl has stronger intermolecular forces than H2O.
Hence, the best argument that compares the intermolecular interactions in water and NaCl is; salt is a solid and water is a liquid, so water has stronger intramolecular forces.
Learn more: brainly.com/question/12108425
Answer:
The answer is 375.54 g of AgBr
Explanation:
Mass (g) = Concentration (mol/L) x volume (L) x Molecular Weight of AgBr (g/mol)
Mass = 2M x 1L x 187.77 g/mol
Mass = 375.54g
Answer:
#1: 0.00144 mmolHCl/mg Sample
#2: 0.00155 mmolHCl/mg Sample
#3: 0.00153 mmolHCl/mg Sample
Explanation:
A antiacid (weak base) will react with the HCl thus:
Antiacid + HCl → Water + Salt.
In the titration of antiacid, the strong acid (HCl) is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.
Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:
Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl
Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl
Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl
The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.
Thus, mmoles HCl /mg OF SAMPLE<em> </em>for each trial is:
#1: 2.178mmol / 1515mg
#2: 2.253mmol / 1452mg
#3: 2.219mmol / 1443mg
<h3>#1: 0.00144 mmolHCl/mg Sample</h3><h3>#2: 0.00155 mmolHCl/mg Sample</h3><h3>#3: 0.00153 mmolHCl/mg Sample</h3>
Answer:
I agree
Explanation:
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