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Tems11 [23]
3 years ago
9

How many atoms of phosphorus are in 4.50 mol of copper(ii) phosphate?

Chemistry
2 answers:
erastova [34]3 years ago
6 0

Answer: 54.2\times 10^{23} atoms

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here copper is having an oxidation state of +2 called as Cu^{2+} cation and phosphate PO_4^{3-} is an anion with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral Cu_3(PO_4)_2.

1 mole of Cu_3(PO_4)_2 contains = 2\times 6.023\times 10^{23}=12.05\times 10^{23} atoms of phosphorous

Thus 4.50 mole of Cu_3(PO_4)_2 contain = \frac{12.05\times 10^{23}}{1}\times 4.50=54.2\times 10^{23} atomsof phosphorous

Thus there are 54.2\times 10^{23} atoms of phosphorous.

stealth61 [152]3 years ago
3 0
<span>Because copper (II) cations have a charge of +2 and phosphate anions have a charge of -3, the formula of anhydrous* copper (II) phosphate is

Cu3(PO4)2

This shows that each formula unit contains 2 atoms of phosphorus. Therefore, the atoms of phosphorus in 4.8 formula units is

4.8 x Avogadro's Number (</span><span>6.022x10*23)</span>, or about 2.9 X 10*24 atoms
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Answer:

3.2 × 10⁻⁸

Explanation:

Let's consider the solution of magnesium carbonate.

MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

We can relate the molar solubility (S) with the solubility product (Ksp) using an ICE chart.

         MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

I                             0                0

C                          +S              +S

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The Ksp is:

Ksp = [Mg²⁺] × [CO₃²⁻] = S × S = S² = (1.8 × 10⁻⁴)² = 3.2 × 10⁻⁸

4 0
4 years ago
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Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
Inessa [10]

Answer:

The answer is "\bold{4.97 \times 10^{-2}}"

Explanation:

Please find the complete question in the attached file.

Equation:

2SO_2+O_2  \leftrightharpoons 2SO_3

at t=0 3.3   \ \ \ \ \ \ \ \ \ \ 0.79

at equilibrium 3.3-p \ \ \ \ \ \ \ \ \ \ 0.79 - \frac{P}{2} \ \ \ \ \ \ \ \ \ \ \ \ P

p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\

     =\frac{0.47^2}{2.83^2\times 0.555}\\\\=4.97 \times 10^{-2}

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Solid cadmium sulfide reacts with an aqueous solution of sulfuric acid . Express your answer as a balanced chemical equation. Id
max2010maxim [7]

Explanation:

When solid cadmium sulfide reacts with an aqueous solution of sulfuric acid then the reaction will be as follows.

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Hence, ionic equation for this reaction is as follows.

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Therefore, net ionic equation for this reaction is as follows.

      CdS(s) + 2H^{+}(aq) \rightarrow Cd^{2+}(aq) + H_{2}S(g)

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