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Lapatulllka [165]
3 years ago
8

What is the solubility in moles/liter for calcium phosphate at 25°C given a Ksp value of 1.2 x 10^-29. Write using scientific no

tation and use 1 or 2 decimal places.
Chemistry
1 answer:
Oksanka [162]3 years ago
6 0

Answer:

Explanation:

Ca₃( PO₄)₂ = 3 Ca⁺² + 2 PO₄⁻³

x                      3x            2 x

Let solubility be x mole per litre

solubility product = (3x )³ x ( 2x )²

= 108 x⁵ = 1.2 x 10⁻²⁹

x⁵ = .0111 x 10⁻²⁹ = .111 x 10⁻³⁰

x = .64 x 10⁻⁶ moles / litre

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Help!!!!!!!!!!!!!!!!!!
skad [1K]

The density of the sample : 0.827 g/L

<h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm , N/m²

V = volume, liter  

n = number of moles  

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)

T = temperature, Kelvin  

n= 1 mol

MW Neon = 20,1797 g/mol

mass of Neon :

\tt mass=mol\times MW\\\\mass=1\times 20,1797 =20.1797~g

The density of the sample :

\tt \rho=\dfrac{m}{V}\\\\\rho=\dfrac{20,1797}{24.4}=0.827~g/L

or We can use the ideal gas formula ta find density :

\tt \rho=\dfrac{P\times MW}{RT}\\\\\rho=\dfrac{1\times 20.1797}{0.082\times 298}\\\\\rho=0.826~g/L

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When adding the measurements 42.1014 g + 190.5 g, the answer has ___ significant figures.
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