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kow [346]
3 years ago
8

How many grams of lithium chloride can be produced from 14.3 g of lithium chlorate when it decomposes into lithium chloride and

oxygen gas?
Chemistry
1 answer:
vampirchik [111]3 years ago
3 0

6.7 grams of lithium chloride will be produced.

<h3><u>Explanation:</u></h3>

Lithium chlorate is LiClO₃ and lithium chloride is LiCl. The reaction is,

2LiClO₃ = 2LiCl +3O₂.

So here, 2 moles of lithium chlorate produces 2 moles of lithium chloride.

Or, one molecule of lithium chlorate will produce one mole of lithium chloride.

Molecular weight of lithium chlorate =6.9 + 35.5 + 16\times3 = 90.4.

So, 14.3 grams of lithium chlorate has 0.16 moles of lithium chlorate.

Thereby, moles of lithium chloride produced is 0.16 moles.

Molecular weight of lithium chloride = 6.9+35.5 = 42.4 grams.

So weight of lithium chloride produced = 42.4 \times 0.16 = 6.7 grams.

Thus, weight of lithium chloride produced will be 6.7 grams.

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Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
Write electron configurations for Gallium, Ga (Z=31), and show the total valence electrons
ivann1987 [24]

<u>Answer:</u> The electronic configuration of gallium is written below and number of valence electrons is 3.

<u>Explanation:</u>

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

Valence electrons are defined as the electrons present in the outermost shell of an atom.

We are given:

An element Gallium having atomic number as 31.

Number of electrons = 31

Electronic configuration of Gallium is: (Z=31):1s^22s^22p^63s^23p^64s^23d^{10}4p^1

This element has 3 electrons in its outermost shell. So, the number of valence electrons is 3

Hence, the electronic configuration of gallium is written below and number of valence electrons is 3.

3 0
3 years ago
Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(
Grace [21]

Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. <u>This statement is true.</u>

<u />

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

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Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so <u>this statement is true.</u>

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

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4 0
3 years ago
What’s the chemical formula for lithium carbonate
nordsb [41]

Answer:

<h3>Being a science geek the ans is <u><em>Li2CO3</em></u></h3>

Explanation:

8 0
3 years ago
Read 2 more answers
A ball is thrown straight up into the air with a speed of 13 m/s. If the ball has
nataly862011 [7]

Answer:

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i.e. E = KE + PE

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At maximum height, velocity=0, thus, KE = 0 and PE = mgh

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or, h = v^2 /2g = 13^2 / (2x9.8) = 8.622 m

Hope this helps.

5 0
3 years ago
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