3.01× 1024 particles are the number of particles are there in 5 grams of sodium carbonate.
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How many particles are there in 5 grams of sodium carbonate?</h3>
There are 6.022 × 1023 particles in one gram of a substance according to Avogadro's number. So when we find out for 5 grams, then we multiply 5 with 6.022 × 1023, we get 3.01 × 1024 particles. For one gram atomic weight of hydrogen, one mole of hydrogen contains 6.022 × 1023 hydrogen atoms.
So we can conclude that 3.01× 1024 particles are the number of particles are there in 5 grams of sodium carbonate.
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Answer:
Mass= 2.77g
Explanation:
Applying
P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28
PV=nRT
NB
Moles(n) = m/M
PV=m/M×RT
m= PVM/RT
Substitute and Simplify
m= (2.09×1.13×28)/(0.082×291)
m= 2.77g
Answer:
130 g of sucrose
Explanation:
Boiling point elevation formula → ΔT = Kb . m
ΔT = Boiling T° solution - Boiling T° pure solvent → 0.39°C
0.39°C = 0.513°C/m . M
m = 0.760 mol/kg → molality = moles of solute / 1kg of solvent
Let's determine the moles of solute → molality . kg
0.760 mol/kg. 0.5 kg = 0.380 moles
If we convert the moles to mass, we'll get the answer
0.380 mol . 342.30 g/mol = 130g
Answer:
A
Explanation:
will group 2 elements gain electrons to bond with non-metals in group 16 in a 2:1 ratio