Colligative properties calculations are used for this type of problem. Calculations are as follows:
ΔT(boiling point) = 101.02 °C - 100.0 °C= 1.02 °C
<span>ΔT(boiling point) = (Kb)m
</span>m = 1.02 °C / 0.512 °C kg / mol
<span>m = 1.99 mol / kg
</span><span>ΔT(freezing point) = (Kf)m
</span>ΔT(freezing point) = 1.86 °C kg / mol (<span>1.99 mol / kg)
</span>ΔT(freezing point) = 3.70 <span>°C
</span>Tf - T = 3.70 <span>°C
T = -3.70 </span><span>°C</span>
Answer:
If Thomson's atomic theory was accurate, the positively charged particles would have gone through the foil.
The balanced chemical equation is written as :
2H3PO4 + 3Ca(OH)2 → 6H2O + Ca3(PO4)2.
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From the above equation we can see that 3 moles of Ca(OH)2 produces 1 mole Ca3(PO4)2 .
( Don't take tension about H3PO4 as it is present in excess )
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Further ,
we are given 78.5 g Ca(OH)2 .
then no of moles Ca(OH)2 given = 78.5/74 = 1.06 moles.
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3 moles Ca(OH)2 produces → 1 mole Ca3(PO4)2.
=> 1 mole Ca(OH)2 produces → 1/3 mole Ca3(PO4)2.
=> 1.06 mole Ca(OH)2 produces → (1/3)×1.06 mole Ca3(PO4)2.
Hence number of moles Ca3(PO4)2 produced = (1/3)×1.06 = 0.353 moles.
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Now gram molecular mass of Ca3(PO4)2 is = 310 g /mole .
hence grams of Ca3(PO4)2 produced
= 0.353× 310 = 109.43 g
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Hope it helps...
Using E=hν where h is Planck's constant and v is the frequency of the photon. In the question above,the wavelenght is given so we can find the frequency of the photon using c=λν. c is a constant =3*10^8 so frequency is equal to
(3*10^8)/0.135*10^-9. Then use ur frequency in the eqn above using h 6.626*10^-34
