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GREYUIT [131]
2 years ago
4

A chemistry student needs of -methyl--pentanone for an experiment. He has available of a w/w solution of -methyl--pentanone in d

iethyl ether. Calculate the mass of solution the student should use
Chemistry
1 answer:
tamaranim1 [39]2 years ago
8 0

The question is incomplete, here is the complete question:

A chemistry student needs of 60.0 g of 4-methyl-2-pentanone for an experiment. He has available 350. g of a 14.0 % w/w solution of 4-methyl-2-pentanone in diethyl ether. Calculate the mass of solution the student should use. If there's not enough solution, press "No solution" button

<u>Answer:</u> There is not enough solution.

<u>Explanation:</u>

We are given:

Mass of 4-methyl-2-pentanone needed = 60.0 gram

14 % w/w 4-methyl-2-pentanone solution

This means that 14 grams of 4-methyl-2-pentanone is present in 100 grams of solution

Applying unitary method:

If 14 grams of 4-methyl-2-pentanone is present in 100 grams of solution

So, 60 grams of 4-methyl-2-pentanone will be present in = \frac{100}{14}\times 60=428.6g

As, the given amount of solution is less than the calculated amount.

Hence, there is not enough solution.

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inna [77]
Molar mass <span>CH2BrCH2Br = 188.0 g/mol

1 mole ---------- 188.0 g
</span>0.500 moles ----- ?

mass = 0.500 * 188.0 / 1

= 94.0 g

Answer C

hope this helps!
4 0
2 years ago
How many moles are there in 8.50 X 1024 molecules of sodium sulfate, Na2SO3?
DiKsa [7]
Hello!

To solve this question, we need to use the Avogadro's Number, which is a constant first discovered by Amadeo Avogadro, an Italian scientist. He discovered that in a mole of a substance, there are 6,02*10²³ molecules. Using this relationship, we apply the following conversion factor:

8,50* 10^{24}molecules* \frac{1 mol Na_2SO_3}{6,02* 10^{23}molecules}=14,12 molesNa_2SO_3

So, 8,50 * 10²⁴ molecules of Na₂SO₃ represent 14,12 moles of Na₂SO₃ 

Have a nice day!
7 0
2 years ago
Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod
son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

In volume V , gram of iodine = V x .12 g

According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

8 0
2 years ago
Be sure to answer all parts. ΔH o f of hydrogen chloride [HCl(g)] is −92.3 kJ/mol. Given the following data, determine the ident
Akimi4 [234]

Answer:

NH₄Cl(s) → NH₃(g) + HCl(g)

ΔH°rxn 74,89 kJ/mol

Explanation:

The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):

<em>(1) </em>¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol

It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).

For the reactions:

<em>(2) </em>N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol

<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

The sum of -(2) + (3) gives:

<em>-(2) </em>2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol

<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)

ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol

Now, the sum of -(2) + (3) + 2×(1)

<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol

<em>2×(1)  </em>H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol

<em>-(2) + (3) + 2×(1) </em>2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)

ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol

The reaction of:

<em>NH₄Cl(s) → NH₃(g) + HCl(g)</em>

<em>Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol</em>

I hope it helps!

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Explanation:
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