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3 years ago

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In many of the steps in the lab, a mixture was centrifuged and then decanted. In general terms, what is accomplished by these st

eps? In what kind of situation would these steps not be able to separate two compounds ?

1 answer:

nirvana33 [79]3 years ago

3 0

Answer:

to separate the compounds by gravity

Explanation:

Centrifuging is the process of separating compound in a liquid mixture by means of gravity and settling. This makes use of the density principle. In the process, the sample is added to the centrifuge. This is then rotated at a certain speed, say 50 rpm. The circular motion creates a force of gravity that pulls the compounds downwards. This then separate the compounds. The heaviest and most dense go down first, then the lighter particles on the top.

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Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

4 0

3 years ago

Chromium is dissolved in sulfuric acid according to the following equation: Cr + H2SO4 ⇒ Cr2 (SO4) 3 + H2

Usimov [2.4K]

Answer:

$\large \boxed{\text{a)188.4 g; b) 98.67 $\, \%$}}$

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 98.08 392.18

2Cr + 3H₂SO₄ ⟶ Cr₂(SO₄)₃ + 3H₂

To solve the stoichiometry problem, you must

Use the molar mass of H₂SO₄ to convert the mass of H₂SO₄ to moles of H₂SO₄
Use the molar ratio to convert moles of H₂SO₄ to moles of Cr₂(SO₄)₃
Use the molar mass of Cr₂(SO₄)₃ to convert moles of Cr₂(SO₄)₃ to mass of Cr₂(SO₄)₃

a) Mass of Cr₂(SO₄)₃

(i) Mass of pure H₂SO₄

$\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}$

(ii) Moles of H₂SO₄

$\text{Moles of H$_{2}$SO}_{4} = \text{141.36 g H$_{2}$SO}_{4} \times \dfrac{\text{1 mol H$_{2}$SO}_{4}}{\text{98.08 g H$_{2}$SO}_{4}} = \text{1.441 mol H$_{2}$SO}_{4}$

(iii) Moles of Cr₂(SO₄)₃

The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ $\text{Moles of Cr$_{2}$(SO$_{4}$)}_{3} = \text{1.441 mol H$_{2}$SO}_{4} \times \dfrac{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}}{\text{3 mol H$_{2}$SO}_{4}} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3}$

(iv) Mass of Cr₂(SO₄)₃ $\text{Mass of Cr$_{2}$(SO$_{4}$)}_{3} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3} \times \dfrac{\text{392.18 g Cr$_{2}$(SO$_{4}$)}_{3}}{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}} = \textbf{188.4 g Cr$_{2}$(SO$_{4}$)}_{3}\\\text{The mass of Cr$_{2}$(SO$_{4}$)$_{3}$ formed is $\large \boxed{\textbf{188.4 g}}$}$

b) Percentage yield

It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.

$\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \% = \dfrac{\text{185.9 g}}{\text{188.4 g}} \times 100 \, \% = \mathbf{98.67 \, \%}\\\\\text{The percentage yield is $\large \boxed{\mathbf{98.67 \, \%}}$}$

7 0

4 years ago

What is the name of the test used to find out the gas produced when calcium reacts with sulfuric acid? *

lys-0071 [83]

Explanation:

Litmus paper is your answer

please mark as brilliant

5 0

3 years ago

Read 2 more answers

Milk is to carton AS. is to pitcher

slega [8]

Lemonade is to a pitcher

6 0

4 years ago

Read 2 more answers

Someone please help!!!!

Blababa [14]

I'm not 100% sure on this one, but I think it's C.

Sorry, I'm not really good with Chemistry.

Hope this helps :)

8 0

3 years ago