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3 years ago

CH4(9) + H2O(g) <> CO(g) + 3 H2(g)

Chemistry

1 answer:

blondinia [14]3 years ago

5 0

Answer:

increasing the temperature decreases the value of the equilibrium constant. Where the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant if u increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again

Explanation:

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What is the most likely way a chemist would be involved in the production of<br> a computer chip?

ZanzabumX [31]

Answer: Computer chips are very complex, maybe there are things insode that chemists are familiar with. Maybe they can give insight about this.

Explanation:

4 0

3 years ago

Calculate the Cal(kcal) in 1 cup of whole milk: 12 g of carbohydrate, 9 g of fat, and 9 g of protein.

laiz [17]

Fat: 1 gram = 9 calories Protein: 1 gram = 4 calories Carbohydrates: 1 gram = 4 calories Alcohol: 1 gram = 7 calories"
So, (12 g carbohydrates)*4 Cal=48 Cal; (9 g fat) *9 Cal=81 Cal; (9 g protein)*4 Cal=36 Cal
48 Cal + 81 Cal + 36 Cal = 165 Cal in one glass of milk

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

7 0

4 years ago

Carbon dioxide and water are produced when ethanol, C2H5OH, is burned in oxygen. The number of moles of CO2 that is produced whe

suter [353]

The number of moles of CO₂ that is produced when burning 6.0 mol of ethanol is 12 mol.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now we have to write the balanced equation

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

We can see that every 1 mole of ethanol we will get 2 mole of CO₂.

So 6.0 mol of ethanol we will get = 6.0 × 2.0

= 12 mol of CO₂

Thus from the above conclusion we can say that The number of moles of CO₂ that is produced when burning 6.0 mol of ethanol is 12 mol.

Learn more about the Balanced Chemical Equation here: brainly.com/question/26694427

#SPJ1

7 0

2 years ago

Bob just filled his car's gas tank with 20 gallons of gasohol, a mixture consisting of 5% ethanol and 95% gasoline. If his car r

WARRIOR [948]

Answer:

C. 10/9 gallons

Explanation:

Given that

V= 20 gallons

Volume of ethanol = 5% of V = 0.05 x 20 = 1 gallons

Volume of the gasoline = 95 % of V = 19 gallons

Lets take x gallons of ethanol is added to achieve optimum performance.

(1+x)/(20+x) = 10 %

(1+x)/(20+x) =0.1

1+ x = 0.1 ( 20+ x)

1+ x = 2 + 0.1 x

1 = 0.9 x

x= 10 /9 gallons

So the option C is correct.

C. 10/9 gallons

5 0

3 years ago