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worty [1.4K]
2 years ago
10

A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate for

m when 10. ml of 0.0010 M Ca(NO3)2 is mixed with 10. ml of 0.00010 M NaF? Ksp for CaF2 = 3.2 x 10-11
Chemistry
1 answer:
laiz [17]2 years ago
5 0

Answer:

CaF2 will not precipitate

Explanation:

Given

Volume of Ca(NO3)2 = 10 ml

Molar concentration of Ca(NO3)2 = 0.001

Volume of NaF = 10 ml

Molar concentration of  NaF  = 0.0001

Ksp for CaF2 = 3.2 * 10^ {-11}

CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2

Moles of calcium ion

= 10 * 0.001\\= 0.01

[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}

Moles of F- ion

= 10 * 0.0001\\= 0.001

[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}

Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}

Q is lesser than Ksp value of CaF2. Hence it will not precipitate

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sleet_krkn [62]

Answer:

Mass of C₂H₄N₂ produced = 3.64 g

Explanation:

The balanced chemical equation for the reaction is given below:

3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)

From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O

Molar masses of the compounds are given below below:

CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol

Comparing the mole ratios of the reacting masses;

CH₄ = 1.65/16 = 0.103

CO₂ = 13.5/44 = 0.307

NH₃ = 2.21/17 = 0.130

converting to whole number ratios by dividing with the smallest ratio

CH₄ = 0.103/0.103 = 1

CO₂ = 0.307/0.103 = 3

NH₃ = 0.130/0.103 = 1.3

Multiplying through with 5

CH₄ = 1 × 5 = 5

CO₂ = 3 × 5 = 15

NH₃ = 1.3 × 5 = 6.5

Therefore, the limiting reactant is NH₃

8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂

Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂

Mass of C₂H₄N₂ produced = 3.64 g

7 0
3 years ago
Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag. Of the two metals, copper
KatRina [158]

Answer:

Answer is given below.

Explanation:

Anode is that electrode where oxidation occurs. Cathode is that electrode where reduction occurs.

In cell representation, half cell present left to salt-bridge notation(\parallel ) is anodic system and another half cell present right to salt-bridge notation(\parallel ) is cathodic system.

So anode is Cu and cathode is Ag.

oxidation: Cu-2e^{-}\rightarrow Cu^{2+}(aq.)

[reduction: Ag^{+}+e^{-}\rightarrow Ag]\times 2

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chemical equation: Cu+2Ag^{+}(aq.)\rightarrow Cu^{2+}(aq.)+2Ag

Oxidizing agent is that species which takes electron from another species. Here Ag^{+}(aq.) takes electron from Cu. Hence Ag^{+}(aq.)  is the oxidizing agent.

Reducing agent is that species which gives electron to another species. Here Cu gives electron to Ag^{+}(aq.) . Hence Cu  is the reducing agent.

6 0
3 years ago
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Explanation:

These elements are rare because:

<u>Helium fuses into the carbon by the combination of three helium nuclei (Z = 2) and one carbon nucleus (Z = 6), therefore bypassing elements with Z= 3, 4 and 5 which are  lithium, beryllium, and boron respectively. Therefore, the fusion processes in cores of the stars do not form these three elements. </u>

7 0
3 years ago
What generalizations can be made about silver halides?
hichkok12 [17]
Halides is the term given to the ions of halogens. Halogens are the second-to-the-last column or period in the periodic table. Examples are chlorine, fluorine, bromine and iodine. Halides are all soluble in water except when combine with silver, lead and mercury. <em>Therefore, the generalization we can make is that silver halides are insoluble in water,</em>
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3 years ago
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Lubov Fominskaja [6]

Answer:

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Explanation:

number of moles = molar concentration x volume in litre

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