Answer:
3g of NaI is needed to prepare 400.0 mL of 0.0500 M NaI solution.
Explanation:
A 0.0500M solution means that is 0.0500 moles of NaI per every 1000 mL of solution. Thus, to prepare 400mL:
Also, the molar mass of NaI is 127g/mol + 23g/mol = 150g/mol
Consequently, multiplying the molar mass of NaI by the moles the necessary mass is obtained:
Parks and open spaces
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Answer:
146 g/mol → option b.
Explanation:
This is a problem about the freezing point depression. The formula for this colligative property is:
ΔT = Kf . m . i
We assume i = 1, so our compound is not electrolytic.
ΔT = Freezing T° of pure solvent - Freezing T° of solution = 1.02 °C
m = molality (mol of solute/kg of solvent)
We convert the grams of solvent (benzene) to kg → 250 g . 1 kg/1000 = 0.250 kg.
We replace → 1.02°C = 5.12°C/mol/kg . mol/ 0.250kg . 1
1.02°C / 5.12 mol/kg/°C = mol/ 0.250kg
0.19922 mol/kg = mol/ 0.250kg
mol = 0.19922 . 0.250kg → 0.0498 mol
molar mass = g/mol → 7.27 g / 0.0498mol = 146 g/mol
When an atom emits a beta
particle from the nucleus, the nucleus only has one more proton and one less
neutron and this will make the atomic mass number remains unchanged while the
atomic number increases by 1.
<span>I hope this answer helps you!</span>
