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katen-ka-za [31]

2 years ago

9

Why is the molar solubility of silver chloride lower in an aqueous solution of silver nitrate than in pure water?

1 answer:

REY [17]2 years ago

5 0

Answer is: because silver cations are common ions in both chemical dissociation in water for this salts, so silver cations shift dissociation od silver chloride to the ledt according <span>Le Chatelier's principle.</span>
Chemical reaction (dissociation) of silver chloride in water:
AgCl(s) → Ag⁺(aq) + NO₃⁻(aq).
Chemical reaction (dissociation) pf silver nitrate in water:
AgNO₃ → Ag⁺(aq) + NO₃⁻(aq).

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4Cr(s)+3O2(g)→2Cr2O3(s) calculate how many grams of the product form when 21.4 g of O2 completely reacts

weqwewe [10]

Answer:

= 67.79 g

Explanation:

The equation for the reaction is;

4Cr(s)+3O2(g)→2Cr2O3(s)

The mass of O2 is 21.4 g, therefore, we find the number of moles of O2;

moles O2 = 21.4 g / 32 g/mol

=0.669 moles

Using mole ratio, we get the moles of Cr2O3;

moles Cr2O3 = 0.669 x 2/3

=0.446 moles

but molar mass of Cr2O3 is 151.99 g/mol

Hence,

The mass Cr2O3 = 0.446 mol x 151.99 g/mol

<u> = 67.79 g </u>

6 0

2 years ago

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Jlenok [28]

Answer:

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Explanation:

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3 years ago

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zloy xaker [14]

Answer:(3)

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6 0

2 years ago

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I. define ionic bonding and describe how it can be formed between sodium and chlorine

Vitek1552 [10]

Math or whatever thing

6 0

3 years ago

What is the empirical formula of the compound that is 25.3% magnesium and 74.7% chlorine

Margarita [4]

Answer: $MgCl_2$

Explanation:

25.3% Mg

74.7% Cl

First step: change % to g

25.3g Mg

74.7g Cl

Second step: calculate g/mol of each compound. You can do this by using the atomic mass.

$25.3gMg(\frac{1mol}{24.30g})=1.04mol$

$74.7gCl(\frac{1mol}{35.45g} )=2.11mol$

Third step: determine the lowest number and divide everything by it. Of the result, extract whole number only.

$Mg=\frac{1.04}{1.04} =1$

$Cl=\frac{2.11}{1.04}=2$

Fourth step: Write each compound with their respective number below.

This empirical formula should be: $MgCl_2$

4 0

3 years ago