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maksim [4K]
2 years ago
14

Why you cannot add 0.3 moles of calcium fluoride directly to 1L of water to make a 0.3 M solution ? please only 5 mins left

Chemistry
1 answer:
Ivenika [448]2 years ago
7 0

Answer:

Because it's not soluble in water, need to be heated acidic solution

Explanation:

Calcium Fluoride Formula: CaF2  Comprises of Ca2+ and F−

H2O = Water H+ O2- OH-

Insoluble in water

In order to dissolve a salt, you have to break apart the ions and hydrate them via a solvent.

Need to read

HSAB concept( Pearson acid-base concept) is an initialism for "hard and soft (Lewis) acids and bases"

So F- is a Halogen which is a Hard Base

And OH- is a Hard base as well

And H- is a soft base

So soft acids react best with soft bases and hard acids react best with hard bases.

Bases don't react with other bases.

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An acid can be defined as a proton donor.
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3 years ago
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A chemist prepares a solution of magnesium chloride MgCl2 by measuring out 49.mg of MgCl2 into a 100.mL volumetric flask and fil
Flura [38]

Answer:  Molarity of Cl^- anions in the chemist's solution is 0.0104 M

Explanation:

Molarity : It is defined as the number of moles of solute present per liter of the solution.

Formula used :

Molarity=\frac{n\times 1000}{V_s}

where,

n= moles of solute

Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.049g}{95g/mol}=5.2\times 10^{-4}moles  

V_s = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

Molarity=\frac{5.2\times 10^{-4}moles\times 1000}{100ml}=5.2\times 10^{-3}mole/L

Therefore, the molarity of solution will be 5.2\times 10^{-3}mole/L

MgCl_2\rightarrow Mg^{2+}+2Cl^-

As 1 mole of MgCl_2 gives 2 moles of Cl^-

Thus 5.2\times 10^{-3}  moles of MgCl_2 gives =\frac{2}{1}\times 5.2\times 10^{-3}=0.0104

Thus the molarity of Cl^- anions in the chemist's solution is 0.0104 M

6 0
3 years ago
Balance the following equations:
lesya [120]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2>

(I) N_{2} + 3H_{2} --> 2NH_{3}

(II) P_{4}  + 5O_{2}  --> 2P_{2} O_{5}

(III) 2NaF + Br_{2}  -->2NaBr + F_{2}

(IV) 2ZnS + 3O_{2} --> 2ZnO + 2SO_2

(V) Pb(NO_3)_2 + 2NaCl --> 2NaNO_3 + PbCl_2

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7 0
2 years ago
The enthalpies of formation of the compounds in the combustion of methane, , are CH4 (g): Hf = –74.6 kJ/mol; CO2 (g): Hf = –393.
algol [13]

Answer:

The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol

Explanation:

The chemical reaction of the combustion of methane is given as follows;

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor

Where:

CH₄ (g): Hf = -74.6 kJ/mol

CO₂ (g): Hf = -393.5 kJ/mol

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Therefore, the combustion of 1 mole of methane releases;

-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol

Hence the combustion of 2 moles of methae will rellease;

2 × -802.54 kJ/mol or 1,605.08 kJ/mol.

8 0
3 years ago
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What is the balanced form of the chemical equation shown below?
Tanzania [10]
B. 1, 1, 1, 2
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7 0
2 years ago
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