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3 years ago

Which of these is an element? Carbon Carbon dioxide Air Water

Chemistry

2 answers:

Natali5045456 [20]3 years ago

8 0

Answer:

carbon

Explanation:

none of the others are on the periodic table

DerKrebs [107]3 years ago

5 0

Answer:

carbon

Explanation:

the rest are mixtures/ compounds.

hope this helps! have a great day!

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EASY Chemical reactions<br> Balance and classify

Marat540 [252]

Answer:

double replacement

synthesis

double replacement

i think this one is decomposition

synthesis

double replacement

single replacement

double replacement

single replacement .....

hopefully i help

6 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

Use the equation:

nadezda [96]

Answer:

37.5 moles of O2 needed

Explanation:

2 moles of C6H6 need 15 moles of O2

5/2 * 15 = 37.5 of O2 needed

8 0

2 years ago

Fe2O3 + Al ---> Fe + Al2O3

PIT_PIT [208]

Answer:

$Fe2O3 + 2Al \rightarrow 2Fe + Al2O3$

Explanation:

Hello!

In this case, when balancing chemical reactions, we must make sure that the atoms of each element are the same at both reactants and products; thus, for the given reaction, we need two iron and aluminum atoms at each side based on their subscripts in the given oxides as shown below:

$Fe2O3 + 2Al \rightarrow 2Fe + Al2O3$

Best regards!

5 0

3 years ago

You have 15 grams of ice at 0°C. How much energy is needed to melt the ice and warm it to 25°C. Ice melts at 0°C. The specific h

azamat

Answer:

d. 1600 calories

Explanation:

The heat of fusion of water, L, is the amount of heat per gram required to melt the ice to water, a process which takes place at a constant temperature of 0 °C. The specific heat of water, c, is the amount of heat required to change the temperature of 1 gram of water by 1 degree Celsius.

We will convert the units of c from Jg⁻¹°C⁻¹ to cal·g⁻¹°C⁻¹ since the answers are provided in calories. The conversion factor is 4.18 J/cal.

(4.18 Jg⁻¹°C⁻¹)(cal/4.18J) = 1 cal·g⁻¹°C⁻¹

First we calculate the heat required to melt the ice, where M is the mass:

Q = ML = (15 g)(80 cal/g) = 1200 cal

Then, we calculate the heat required to raise the temperature of water from 0 °C to 25 °C.

Q = mcΔt = (15 g)(1 cal·g⁻¹°C⁻¹)(25 °C - 0 °C) = 380 cal

The answer is rounded so that there are two significant figures

The total heat required for this process is (1200 cal + 380 cal) = 1580 cal

The rounded answer is 1600 calories.

3 0

3 years ago