Yes because usually it’s regardless of how small or big it is
After the first half-life, 1/2 is unchanged. After the 2nd one, 1/4 remains. After the 4th one, 1/16 remains. So you have 4 half-lives in 48 hours, the half-life is thus 12 hours.
False evaporation would do the trick
The required moles of AgBr precipitate produced by given moles of silver nitrate is 0.0123.
<h3>How do we calculate moles from molarity?</h3>
Molarity of any solution is define as the moles of solute present in per liter of the solution and it will be represented as:
M = n/V
Given that, molarity of AgNO₃ = 0.250M
Volume of AgNO₃ = 49.5mL = 0.0495L
Moles of AgNO₃ = (0.25)(0.0495) = 0.0123mol
Given chemical reaction is:
2AgNO₃(aq) + CaBr(aq) → 2AgBr(s) + Ca(NO₃)₂(aq)
As it is mention that CaBr is present in excess quantity and AgNO₃ is the limiting reagent so the formation of precipitate will depend on the AgNO₃.
From the stoichiometry of the reaction, it is clear that:
2 moles of AgNO₃ = produces 2 moles of AgBr
0.0123 moles of AgNO₃ = produces 2/2×0.0123=0.0123 moles of AgBr
Hence 0.0123 is the required moles of precipitate.
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142 g ---------------- 6.02 x 10²³ molecules
0.920g g ------------ ( molecules )
molecules = 0.920 x ( 6.02 x 10²³ ) / 142
molecules = 5.53 x 10²³ / 142
= 3.89 x 10²¹ molecules
1 molecule P2O5 -------------------------- 7 atoms
3.89 x 10²¹ molecules -------------------- ( atoms )
atoms = ( 3.89 x 10²¹) x 7 / 1
atoms = <span>2.72 x 10²² atoms of P2O5
Hope this helps!!!
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