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Alja [10]
3 years ago
11

Q 3.6: calculate the number of po43- ions present in 23.7 g of ca3(po4)2.

Chemistry
2 answers:
malfutka [58]3 years ago
6 0
Moles of ca3(po4)2 = 23.7 / 310.17 =  0.076

moles of (PO4)3- = 0.076 x 2 = 0.152

now, no. of ions = 0.152 x 6.022 x 10^{23}

                          =  9.2 x 10^{22}
disa [49]3 years ago
6 0

23.7 g of Ca₃(PO₄)₎₂ contains 9.20 x 10^22 ions of PO₄³⁻  

23.7 g of Ca₃(PO₄)₎₂ contains <u>0.1528 moles of PO₄³⁻ </u>which is equivalent to <u>9.20 x 10^22 ions of PO₄³⁻</u>

<h2>Further Explanation:</h2><h3>An Atom</h3>
  • An atom is the smallest particle of an element that can take part in a chemical reaction. For example, sodium atom, Na.  
<h3>A Molecule</h3>
  • A molecule is a substance that is made up of one or more atoms. The atoms in a molecule may be similar or different making a compound.  
<h3>A compound  </h3>
  • A compound is a substances that contains two or more different atoms that are bonded together. When the atoms are similar the substance is known as a molecule, therefore not all molecules are compounds.
  • Ca₃(PO₄)₎₂  is an example of a compound that contains three calcium ions and two Phosphate ions.

Therefore;

Ca₃(PO₄)₎₂ ionizes to give out 3 Ca²⁺ ions and 2 PO₄³⁻ ions

Ca₃(PO₄)₎₂ → 3Ca²⁺ + 2PO₄³⁻

Required to determine the number of ions of PO₄³⁻ ions

<h3>Step 1: Moles of Ca₃(PO₄)₎₂</h3>

Number of moles = Mass /Molar mass

Molar mass of Ca₃(PO₄)₎₂ = 310.1767 g/mol

Thus;

Moles = 23.7 g/310.1767 g/mol

          = 0.0764 moles

<h3>Step 3: Moles of PO₄³⁻ ions</h3>

1 mole of  Ca₃(PO₄)₎₂ ionizes to give 2 moles PO₄³⁻ ions

Therefore;

0.0764 moles will produce;

 = 0.0764 moles x 2

 = 0.0152 moles PO₄³⁻ ions

<h3>Step 3: Number of PO₄³⁻ ions</h3>

1 mole of PO₄³⁻ ions contains 6.022 x 10^23 ions

Therefore;

0.01528 moles PO₄³⁻ ions contains;

= 0.01528 x 6.022 x 10^23 ions

= 9.20 x 10^22 ions

Keywords: Atoms, ions, Moles, molar mass

<h3>Learn more about; </h3>
  • Atom: brainly.com/question/1841136
  • Molecules brainly.com/question/1841136
  • Compounds brainly.com/question/1841136
  • Avogadro’s constant: brainly.com/question/10466480
  • Molecular mass: brainly.com/question/10466480

Level: High school

Subject: Chemistry

Topic: Moles

Sub-topic: Avogadro's constant

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A 250 mL sample of gas is collected over water at 35°C and at a total pressure of 735 mm Hg. If the vapor pressure of water at 3
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Answer:

The volume of the gas sample at standard pressure is <u>819.5ml</u>

Explanation:

Solution Given:

let volume be V and temperature be T and pressure be P.

V_1=250ml

V_2=?

P_{total}=735 mmhg

1 torr= 1 mmhg

42.2 torr=42.2 mmhg

so,

P_{water}=42.2mmhg

T_1=35°C=35+273=308 K

Now

firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

P_{gas}=P_{total}-P_{water}

=735-42.2=692.8 mmhg

Now

By using combined gas law equation:

\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}

V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1

V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1

Here P_2 \:and\: T_2 are standard pressure and temperature respectively.

we have

P_2=750mmhg \:and\: T_2=273K

Substituting value, we get

V_2=\frac{692.8}{750}*\frac{273}{308} *250

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Calculate the concentration of hydronium and hydroxide ions in a 0.050 M solution of nitric acid.
Mazyrski [523]

Answer:

[H₃O⁺] = 0.05 M & [OH⁻] = 2.0 x 10⁻¹³.

Explanation:

  • HNO₃ is completely ionized in water as:

<em>HNO₃ + H₂O → H₃O⁺ + NO₃⁻.</em>

  • The concentration of hydronium ion is equal to the concentration of HNO₃:

[H₃O⁺] = 0.05 M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] </em>= 10⁻¹⁴/0.05 = <em>2.0 x 10⁻¹³.</em>

3 0
2 years ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
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  • Final temperature of water is 23.68 °C
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We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

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<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
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