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zheka24 [161]
2 years ago
6

A student used 6.34 x 10-3 kg of t-butanol in the synthesis of BHT. Assuming no other substances are limiting, how many grams of

BHT can be formed? 9.42 g BHT 6.11 g BHT 18.8 g BHT 12.9 g BHT
Chemistry
2 answers:
aalyn [17]2 years ago
6 0

<u>Answer:</u> The mass of BHT that can be formed is 18.8 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of t-butanol = 6.34\times 10^{-3}kg=6.34g      (Conversion factor: 1 kg = 1000 g)

Molar mass of t-butanol = 74 g/mol

Putting values in equation 1, we get:

\text{Moles of t-butanol}=\frac{6.34g}{74g/mol}=0.086mol

The chemical equation for the formation of BHT from t-butanol follows:

\text{t-butanol + p-cresol}\rightarrow \text{BHT}

By Stoichiometry of the reaction:

1 mole of t-butanol produces 1 mole of BHT

So, 0.086 moles of t-butanol will produce = \frac{1}{1}\times 0.086=0.086mol of BHT

Now, calculating the mass of BHT from equation 1, we get:

Molar mass of BHT = 220 g/mol

Moles of BHT = 0.086 moles

Putting values in equation 1, we get:

0.086mol=\frac{\text{Mass of BHT}}{220g/mol}\\\\\text{Mass of BHT}=(0.086mol\times 220g/mol)=18.8g

Hence, the mass of BHT that can be formed is 18.8 grams

Masteriza [31]2 years ago
6 0

Answer:

yeet

Explanation:

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Answer : BaSO_{4} will be the precipitate which will be formed.


Explanation : When all the three solutions namely; NaSO_{4}  + Ba(NO_{3})_{2}  + NH_{4} ClO_{4} are mixed together a white precipitate of BaSO_{4} is formed as a product in the solution along with the soluble by product of Ammonium nitrate which is NH_{4} NO_{3} 

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A state property X has a value 89.6 units. It undergoes the following changes, first increase by
yanalaym [24]

Answer:

B) -4.1 units

Explanation:

According to this question, a state property X has a value 89.6 units. It undergoes the certain changes as follows:

- first increase by 3.6 units

- then increase by another 18.7 units

- then decrease by 12.2 units

- and finally attains a value of 85.5 units

This can be mathematically represented by 89.6 - {3.6 + 18.7 - 12.2 - x) = 85.5

To get x, we say;

89.6 + 3.6 = 93.2

93.2 + 18.7 = 111.9

111.9 - 12.2 = 99.7

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2 years ago
A sample of a certain barium chloride hydrate, BaCl2.nH2O, has a mass of 7.62 g. When this sample is heated in a crucible, it is
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Answer:

BaCl2.2H2O

Explanation:

Data obtained from the question include:

Mass of barium chloride hydrate (BaCl2.nH2O) = 7.62g

Mass of anhydrous barium chloride (BaCl2) = 6.48g

Next, we shall determine the mass of water in the barium chloride hydrate (BaCl2.nH2O). This can be obtained as follow:

Mass of water (H2O) = Mass of barium chloride hydrate (BaCl2.nH2O) – Mass of anhydrous barium chloride (BaCl2)

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Next, we shall write the balanced equation for the reaction. This is given below:

BaCl2.nH2O —> BaCl2 + nH2O

Next, we shall determine the masses of BaCl2.nH2O that decomposed and the mass H2O produced from the balanced equation.

Molar mass of BaCl2.nH2O = 137 + (35.5x2) + n[(2x1) + 16]

= 137 + 71 + n[2 + 16]

= (208 + 18n) g/mol

Mass of BaCl2.nH2O from the balanced equation = 1 x (208 + 18n) = (208 + 18n) g

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Mass of H2O from the balanced equation = n x 18 = 18n g

Summary:

From the balanced equation above,

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18n x 7.62 = 1.14(208 + 18n)

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Collect like terms

137.16n – 20.52n = 237.12

116.64n = 237.12

Divide both side by the coefficient of n i.e 116.64

n = 237.12/116.64

n = 2

Therefore the formula for the hydrate, BaCl2.nH2O is BaCl2.xH2O.

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