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3 years ago

Write the equilibrium constant: Pb3(PO4)2(s) = 3Pb2+ (aq) + 2PO2 (aq)

Chemistry

1 answer:

kicyunya [14]3 years ago

7 0

Answer:

Kc = [Pb²⁺]³.[PO₄³⁻]²

Explanation:

Let's consider the following reaction at equilibrium.

Pb₃(PO₄)₂(s) ⇄ 3 Pb²⁺(aq) + 2 PO₄³⁻(aq)

The concentration equilibrium constant is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

Kc = [Pb²⁺]³.[PO₄³⁻]²

This equilibrium constant is known as the solubility product of Pb₃(PO₄)₂.

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Explanation:

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3 years ago

Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope

morpeh [17]

Answer:

For a: The isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b: The isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c: The isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

Explanation:

The isotopic representation of an atom is: $_Z^A\textrm{X}$

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

For a:

We are given:

Number of neutrons = 78

Atomic number of iodine = 53 = Number of protons

Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b:

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c:

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

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3 years ago

Name the following ionic compounds. Use the periodic table, the table of common polyatomic ions, and

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Answer: aluminum oxide

Explanation:

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3 years ago

n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains

Nitella [24]

Answer : The metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of unknown metal = ?

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$m_1$ = mass of unknown metal = 150 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature of water = $34.3^oC$

$T_1$ = initial temperature of unknown metal = $150.0^oC$

$T_2$ = initial temperature of water = $25.0^oC$

Now put all the given values in the above formula, we get

$150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC$

$c_1=0.449J/g^oC$

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

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4 years ago