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patriot [66]
2 years ago
10

How many g i2 should be added to 750g ccl4 to prepare a 0.200m solution?

Chemistry
2 answers:
vesna_86 [32]2 years ago
6 0
<span>There are a number of ways to express concentration of a solution. This includes molality. Molality is expressed as the number of moles of solute per mass of the solvent. We calculate as follows:

0.200 mol I2 / kg CCl4 ( .750 kg CCl4 ) ( 253.809 g I2 / mol I2) = 38.07 g I2 needed

Hope this helps.

 </span>
tamaranim1 [39]2 years ago
3 0

Answer:

Mass of I2 required = 38.1 g

Explanation:

<u>Given:</u>

Mass of CCl4 = 750 g = 0.750 kg

Molality of the solution = 0.200 m

<u>To determine:</u>

Mass of I2

<u>Formula:</u>

Molality of a solution is expressed as the ratio of the amount of solute per kg of solvent

Molality = \frac{moles\ of\ solute}{kg\ solvent} \\

Here the solute is I2 and solvent is CCl4

Moles\ of\ I2 = molality*mass\ solvent = 0.200*0.750= 0.150 moles\\\\Molar\  mass\  of\  I2 = 253.8 g/mol\\\\Mass\ of\ I2 = moles*molar\ mass = 0.150*253.8 = 38.1 g

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Sergeeva-Olga [200]
The answer is 3.39 mol.

<span>Avogadro's number is the number of molecules in 1 mol of substance.
</span><span>6.02 × 10²³ molecules per 1 mol.
</span>2.04 × 10²⁴<span> molecules per x.

</span>6.02 × 10²³ molecules : 1 mol = 2.04 × 10²⁴ molecules : x
x = 2.04 × 10²⁴ molecules * 1 mol : 6.02 × 10²³ molecules
x = 2.04/ 6.02 × 10²⁴⁻²³ mol
x = 0.339 × 10 mol
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8 0
2 years ago
A 0.941 gram piece of magnesium metal is heated and reacts with oxygen. the resulting oxide weighed 1.560 grams. determine the p
Whitepunk [10]

Given:

magnesium = 0.941 gram piece

Magnesium oxide= 1.560 grams

Formula:

(magnesium / magnesium oxide) x 100 = % Mg

100% - % Mg = percent composition of each element

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(0.941g Mg) / (1.560g MgO)

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39.7% is the percentage composition of each element.

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Standard Hydrogen Electrode consists of platinum wire fused in a glass tube and a platinum plate coated with finely divided plat
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To increase surface area of the platinum electrode which results in superior quality and action of the electrodes as opposed to normal platinum electrodes.

Explanation:

Platinization of Platinum is the process of covering platinum electrode with a layer of platinum black. Platinum black is a finally divided form of platinum, optimized for catalysing the addition of hydrogen to unsaturated organic compound. This increases the surface area of the platinum electrodes and therefore exhibits action superior to that of normal electrodes.

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Answer:

429.4 kJ are absorbed in the endothermic reaction.

Explanation:

The balanced chemical equation tells us that 1168 kJ of heat are absorbed in the reaction when 4 mol of NH₃ (g) react with 5 mol O₂ (g).

So what we need is to calculates how many moles represent 25 g NH₃(g) and calculate the heat absorbed. (NH₃ is the limiting reagent)

Molar Mass NH₃  = 17.03 g/mol

mol NH₃ = 25.00 g/ 17.03 g/mol = 1.47 mol

1168 kJ /4 mol NH₃  x 1.47 mol  NH₃ =  429.4 kJ

6 0
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As noted in Assignment 1.4.1., alkylation of benzene with 1-chlorobutane in the presence of AlCl3 gives both butylbenzene and (1
Sunny_sXe [5.5K]

Answer:

Use the Bromotriflouride catalyst, BF₃

Explanation:

The BF₃ is most likely to yield less desired side products. The effect lies in the reaction mechanism.

BF₃ is a Lewis acid. Its role is to promote the ionization of the HF. This is achieved through the electrophilic mechanism. The reaction mechanism is as follows:

2 - methylpropene + H-F-BF₃ → H-F + H₃C + benzene

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6 0
2 years ago
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