The mass of lime that can be produced from 4.510 Kg of limestone is calculated as below
calculate the moles of CaCO3 used
that is moles =mass/molar mass
convert Kg to g = 4.510 x1000 =4510g
= 4510 / 100 =45.10 moles
CaCO3 = CaO +O2
by use of mole ratio between CaCO3 to CaO (1:1) the moles of CaO is also= 45.10 moles
mass of CaO = moles x molar mass
45.10 x56 = 2525.6 g of CaO
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Answer:
3) NaCl.
Explanation:
<em>∵ ΔTf = iKf.m</em>
where, <em>i</em> is the van 't Hoff factor.
<em>Kf </em>is the molal depression freezing constant.
<em>m</em> is the molality of the solute.
<em>The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. </em>
<em></em>
- For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.
<em>So, for sugar: i = 1.</em>
<em>∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.</em>
<em></em>
- For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.
<em>So, i for NaCl = 2.</em>
<em>∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.</em>
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<em>So, the right choice is: 3) NaCl.</em>
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Answer:
It's not correct. For balancing, we need to put the coefficients in the molecule, not in the athom. Because if you do this, you're creating another molecule, instead of a balacing, for which the reaction may not happen - but anyway, it would be another reaction.
The correct balacing is:
2NaOH + 1H2S → 1Na2S + 2H2O
Explanation:
Look: Na2OH does not even exist. OH has only one free link, so he can't - in normal conditions - make another one with any athom. That's why we should write 2NaOH instead of Na2OH. The first means "2 mols of NaOH".