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3 years ago

Single, double, or triple covalent bonds in a carbon atom are formed from.

Chemistry

1 answer:

Alex73 [517]3 years ago

3 0

Answer:

Carbon-Carbon Bonds

Carbon can form single, double, or even triple bonds with other carbon atoms. In a single bond, two carbon atoms share one pair of electrons. In a double bond, they share two pairs of electrons, and in a triple bond they share three pairs of electrons

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How much heat (in joules) is required to completely melt 10.00 g of ice starting at 0 oc?

Yuri [45]

/////// 400 cal ///////

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3 years ago

How many grams of aluminum chloride can be produced from 7.32g of ammonium perchlorate?

Alla [95]

Answer:

2.77g of $AlCl_{3}$ can be produced.

Explanation:

Use stoichiometry.

Balanced equation: $3Al + 3NH_{4} ClO_{4}$ ⇒ $Al_{2} O_{3} + AlCl_{3} +3NO + 6H_{2} O$

$NH_{4} ClO_{4} : AlCl_{3}$

3 : 1 ** $n=\frac{m}{M}$

0.0623** : $x$ $n=\frac{7.32}{117.49}$ = 0.0623

$3x=0.0623\\x=0.020767$

$n=\frac{m}{M}$

$m=nM$

m= (0.020767)(133.34)

m= 2.77g

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2 years ago

Name two different signs of a chemical reaction

olasank [31]

A change in temperature or color are two signs that a chemical reaction has taken place.

7 0

4 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 $\large \boxed{\mathbf{98.67 \, \%}}$}$

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4 years ago

Which of the following is an indication of a chemical reaction?

vovikov84 [41]

Answer:

All of the above

Explanation:

Their all chemical change

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3 years ago