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Olenka [21]
3 years ago
9

Silver nitrate solution reacts with lithium chloride solution. Identify the precipitate or precipitates formed.

Chemistry
2 answers:
alina1380 [7]3 years ago
8 0
AgCl is the precipitate
antiseptic1488 [7]3 years ago
5 0

Answer:  AgCl

Explanation:  

When silver nitrate solution reacts with lithium chloride, 2 products are formed which are Silver Chloride (AgCl) and Lithium Nitrate (LiNO3).

But only one is maintained as precipitate and the other one gets solubilize in the solution.

Thus LiNO3 gets solubilize in the solution and AgCl remains as precipitate which does not get dissolve in the solution.

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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
1) How many moles are in 4.0x10^24 atoms?
-Dominant- [34]

Answer:

<h2>6.64 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{4 \times  {10}^{24} }{6.02 \times  {10}^{23} }   \\  = 6.644518...

We have the final answer as

<h3>6.64 moles</h3>

Hope this helps you

3 0
3 years ago
The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell
marysya [2.9K]

Answer:

\boxed{\rm \text{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

Explanation:

The half-cell reduction potentials are

Ag⁺(aq) +   e⁻ ⇌ Ag(s)     E° =  0.7996 V

Fe²⁺(aq) + 2e⁻ ⇌ Fe(s)     E° = -0.447    V

To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.

The anode is the electrode at which oxidation occurs.

The equation for the oxidation half-reaction is

\boxed{\rm \textbf{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

4 0
3 years ago
Which element in Period 3 has the largest atomic radius?
horsena [70]

Answer:

sodium

Explanation:

6 0
2 years ago
A clear liquid in an open container is allowed to evaporate. After three days, a solid is left in the container. Was the clear
Mashutka [201]

The liquid did not chemically bond after 3 days, therefor it is a mixture.

Hope this helps!

7 0
2 years ago
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