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2 years ago

Using dimensional analysis convert 238.0 cm3 to L

Chemistry

1 answer:

Bezzdna [24]2 years ago

7 0

Answer:

= 0.238 L

Explanation:

From the given information:

To convert cm³ to L(liter), we utilize the advent of a unit fraction that comprises a liter in the upper part of the fraction(numerator) and a cubic centimeter in the lower part of the fraction(denominator)

Recall that:

1000 cm³ = 1 liter

∴

238.0 cm³ = $\dfrac{238.0 \ cm^3}{1} \times \dfrac{1 \ L}{1000 \ cm^3}$

= 0.238 L

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Answer:

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Explanation:

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3 years ago

Write the complete balanced equation for the reaction between iron (III) oxide (Fe2O3) and water (H2O). You do not need to make

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Fe2O3+3H2O->2Fe(OH)3

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3 years ago

Consider the reaction. 2 Pb ( s ) + O 2 ( g ) ⟶ 2 PbO ( s ) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of

Art [367]

Answer : The percent yield of the reaction is, 75.6 %

Solution : Given,

Mass of Pb = 451.4 g

Molar mass of Pb = 207 g/mole

Molar mass of PbO = 223 g/mole

First we have to calculate the moles of Pb.

$\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles$

Now we have to calculate the moles of $PbO$

The balanced chemical reaction is,

$2Pb(s)+O_2(g)\rightarrow 2PbO(s)$

From the reaction, we conclude that

As, 2 mole of $Pb$ react to give 2 mole of $PbO$

So, 2.18 mole of $Pb$ react to give 2.18 mole of $PbO$

Now we have to calculate the mass of $PbO$

$\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO$

$\text{ Mass of }PbO=(2.18moles)\times (223g/mole)=486.1g$

Theoretical yield of $PbO$ = 486.1 g

Experimental yield of $PbO$ = 367.5 g

Now we have to calculate the percent yield of the reaction.

$\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}\times 100$

$\% \text{ yield of the reaction}=\frac{367.5g}{486.1g}\times 100=75.6\%$

Therefore, the percent yield of the reaction is, 75.6 %

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3 years ago

What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +

nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ + S₂O₃²⁻ + H₂O → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻ Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

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3 years ago

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