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4 years ago

Identify the difference between natural and artificial dyes.

2 answers:

6 0

Answer: natural dyes is dye that it is anything that grow on earth that haven't been possessed and artificial dye mean that it is good for something in your body or something else

Explanation:i search it up

Art [367]4 years ago

4 0

Answer:

Natural dyes, although pure from chemicals, are often times more expensive than chemical dyes.

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Hello, <br><br> So I was wondering if this is correct... Is it?

Sonja [21]

Looks correct but the second to last I would of put abiotic and biotic factors but I don’t know what’s right for you

4 0

3 years ago

Read 2 more answers

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

Timothy, a cashier, is given a twenty-dollar bill at his store. He is suspicious of the customer and wants to make sure the curr

Mice21 [21]

"<span>There is no change in the color of the bill's ink" is the one clue among the following choices given in the question that the currency is counterfeit. The correct option among all the options that are given in the question is the fourth option or the last option. I hope that the answer has come to your desired help.</span>

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4 years ago

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When molecules collide with the walls of their container, is exerted

evablogger [386]

I believe your answer would be friction...

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4 years ago

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Compared to the energy of an electron in the second shell of an atom of sulfur, the energy of an electron in the

kap26 [50]

Answer:

First shell is lower

Explanation:

The electron has to absorb energy to excite and go to higher energy levels/ shells and should release energy to go to lower energy levels/ shells. This means higher the shell , more energy is required and vice versa

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2 years ago