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4 years ago

The yeast acted as a catalyst. Explain the purpose of a catalyst in a reaction

Chemistry

1 answer:

Anna71 [15]4 years ago

5 0

Answer:

The catalyst speeds up the chemical reaction.

Explanation:

The process in which chemical reaction rate increases by addition of substance is termed as a catalyst. The chemical reaction happens faster when a catalyst is present because it gives an alternative reaction pathway with lower activation energy.

The catalyst usually forms a temporary intermediate that helps in regenerating the original catalyst in the cyclic process. Yeast when added in a mixture work as a catalyst and helps with chemical reaction.

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Both the brain and the spinal cord are protected by _____.

patriot [66]

It is a watery fluid called <span>cerebrospinal fluid. Answer C :)
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3 years ago

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This balanced chemical equation represents a chemical reaction: 6no + 4nh3 → 5n2 + 6h2o what volume of nh3 gas, at standard temp

Liula [17]

The answer is: volume of ammonia gas is 7.4 L.

Chemical reaction: 6NO + 4NH₃ → 5N₂ + 6H₂O.

m(NO) = 15 g; mass of nitrogen(II) oxide.

M(NO) = 30 g/mol; molar mass of nitrogen(II) oxide.

V(NH₃) = ?

n(NO) = 15 g ÷ 30 g/mol.

n(NO) = 0.5 mol; amount of nitrogen(II) oxide.

From chemical reaction: n(NO) : n(NH₃) = 6 : 4.

0.5 mol : n(NH₃) = 6 : 4.

n(NH₃) = 0.33 mol; amount of ammonia.

Vm = 22.4 L/mol; molar volume at STP.

V(NH₃) = 0.33 mol · 22.4 L/mol..

V(NH₃) = 7.4 L.

3 0

4 years ago

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Hydrogen sulfide (H2S) is a gas that smells like rotten eggs. It can be produced by bacteria in your mouth and contributes to ba

Doss [256]

Answer:

See explanation.

Explanation:

Hello!

In this case, we consider the questions:

a. Ideal gas at:

i. 273.15 K and 22.414 L.

ii. 500 K and 100 cm³.

b. Van der Waals gas at:

i. 273.15 K and 22.414 L.

ii. 500 K and 100 cm³.

Thus, we define the ideal gas equation and the van der Waals one as shown below:

$P^{id}=\frac{nRT}{V}\\\\ P^{vdW}=\frac{RT}{v-b}-\frac{a}{v^2}$

Whereas b and a for hydrogen sulfide are 0.0434 L/mol and 4.484 L²*atm / mol² respectively, therefore, we proceed as follows:

i. 273.15 K and 22.414 L.

$P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L}=1 .00 atm$

ii. 500 K and 100 cm³ (0.1 L).

$P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*500K}{0.100L}=410.3 atm$

i. 273.15 K and 22.414 L: in this case, v = 22.414 L / 1.00 mol = 22.414 L/mol

$P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(22.414L/mol)^2}=0.993atm$

ii. 500 K and 100 cm³: in this case, v = 0.1 L / 1.00 mol = 0.100 L/mol

$P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*500K}{0.100L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(0.100L/mol)^2}=276.5atm$

Whereas we can see a significant difference when the gas is at 500 K and occupy a volume of 0.100 L.

Best regards!

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3 years ago

How many significant figures are in 382.90?

Leviafan [203]

it has three significant figures

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4 years ago

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Suppose you seal 30 g of ag2o(s) in a 1.00 liter evacuated vessel, and heat it to 450 k. what pressure (of o2) would develop in

kodGreya [7K]

<span>We would use the ideal gas equation. PV = NRT
No of moles of ag2o = 30/ 231.735 = 0.12944
The chemical reaction is given by
2 Ag2O = 4 Ag + O2
2* 0.12944 of Ag2O requires 0.12944 of O2
So we have P * 1 = 0.258 * 8.2 * 450
P = 952.02 atm</span>

3 0

4 years ago