What is the numerical value of gold silver and platinum?

1 answer:

Klio2033 [76]2 years ago

3 0

Platinum:

850 Platinum: 85% platinum
950 Platinum: 95% platinum
900 Platinum: 90% platinum

Gold:

24 karat: 99.9% gold; the purest form.
22 karat: 91.7% gold
18 karat: 75% gold
14 karat: 58.3% gold
10 karat: 41.7% gold

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When 15.3 g of sodium nitrate, NaNO3,was dissolved in water in a calorimeter, the temperature fell from 25.00oC to 21.56oC. If t

satela [25.4K]

Answer:

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

Explanation:

Heat lost by solution ad calorimeter = Q

Heat capacity of solution ad calorimeter = C = 1071 J/°C

Change in temperature = ΔT = 21.56°C - 25.00°C = -3.44°C

$Q=C\times Delta T$

$Q=1071 J/^oC\times (-3.44^oC)=-3,684.24 J$

Heat gained by sodium nitrate = -Q = -(-3,684.24 J)=3,684.24 J

Moles of sodium nitrate = $\frac{15.3 g}{85 g/mol}=0.18 mol$

When 0.18 mole of sodium nitrate was dissolved in water 3,684.24 joulesof heat was absorbed by it.

Then heat absorbed by 1 mole of sodium nitrate :

$\frac{3,684.24 J}{0.18}=20,468 J=20.468 kJ$

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20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

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3 years ago

Oxana [17]

Answer is: f<span>ormula for the hydrated compound is CuSO</span>₄·3H₂O.
ω(H₂O) = 25,3% = 0,253.
ω(CuSO₄) = 100% - 25,3%.
ω(CuSO₄) = 74,7% = 0,747.
ω(H₂O) : M(H₂O) = ω(CuSO₄) : M(CuSO₄).
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M(H₂O) = (0,253 · 159,6 g/mol) ÷ 0,747.
M(H₂O) = 54 g/mol.
N(H₂O) = 54 g/mol ÷ 18 g/mol.
N(H₂O) = 3.

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