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Allisa [31]
3 years ago
6

In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.

0 cm^3. If the sphere were broken down into eight spheres each having a volume of 1.25 cm^3, and the reaction is run a second time, which of the following accurately characterizes the second run?
Choose all that apply.
A. The second run will be faster.
B. The second run will be slower.
C. The second run will have the same rate as the first.
D. The second run has twice the surface area.
E. The second run has eight times the surface area.
F. The second run has 10 times the surface area.
Chemistry
1 answer:
mamaluj [8]3 years ago
8 0

Answer:

D

Explanation:

We know that the

reaction catalyzing power of a catalyst ∝ surface area exposed by it

Given

volume V1= 10 cm^3

⇒\frac{4}{3} \pi r^3= 10

hence r= 1.545 cm

also, surface area S1= 4\pi r^2

now when the sphere is broken down into 8 smaller spheres

S2= 8×4πr'^2

now, equating V1 and V2 ( as the volume must remain same )

\frac{4}{3}\pi r^3=8\times\frac{4}{3} \pi r'^3

and solving we get

r'= r/2

therefore, S2=8\times4\pi\frac{r}{2}^2

S2=2\times4\pi r^2

S2= 2S1

hence the correct answer is

. The second run has twice the surface area.

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What is the total volume of gaseous products formed when 116 liters of butane (C4H10) react completely according to the followin
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<u>Answer:</u> The total volume of the gaseous products is 1044.29 L

<u>Explanation:</u>

We are given:

Volume of butane = 116 L

At STP:

22.4 L of volume is occupied by 1 mole of a gas

So, 116 L of volume will be occupied by = \frac{1}{22.4}\times 116=5.18mol of butane

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

  • <u>For carbon dioxide:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 5.18 moles of butane will produce = \frac{8}{2}\times 5.18=20.72mol of carbon dioxide

Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L

  • <u>For water vapor:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

So, 5.18 moles of butane will produce = \frac{10}{2}\times 5.18=25.9mol of water vapor

Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L

Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L

Hence, the total volume of the gaseous products is 1044.29 L

3 0
2 years ago
The process of fermentation involves chemical reactions that convert solid glucose (C6H12O6) into aqueous ethanol and carbon dio
e-lub [12.9K]

Answer:

C6H12O6 —> 2C2H5OH + 2CO2

Explanation:

The equation for the reaction is given below:

C6H12O6 —> C2H5OH + CO2

We can balance the equation above as follow:

There are 12 atoms of H on the left side and 6 atoms of the right side. It can be balance by putting 2 in front of C2H5OH as shown below:

C6H12O6 —> 2C2H5OH + CO2

There are 6 atoms of C on the left side and 5 atoms on the right side. It can be balance by putting 2 in front of CO2 as shown below:

C6H12O6 —> 2C2H5OH + 2CO2

Now the equation is balanced.

4 0
3 years ago
Answer Quick
kow [346]

Answer:

20.9%

Explanation:

  • The percentage by mass of solution is given by dividing the mass of solute in grams by the mass of solution in grams then multiplying it by 100%.

% Mass of solution = mass of solute/mass of solution × 100%

                               = (27.0 g/ 129.0 g) × 100%

                               = 20.93%

                               = 20.9%

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