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djyliett [7]
3 years ago
7

What is the theoretical yield of so3 produced by the quantities described in part a?\?

Chemistry
1 answer:
d1i1m1o1n [39]3 years ago
8 0
The problem here is incomplete. Luckily, I found a similar problem from another website which is shown in the attached picture. So, the useful information here are the stoichiometric coefficients of the reaction and the initial amount of 7.49 g of S. The solution is as follows:

Molar mass of S = 32.06 g/mol
Molar mass of SO₃ = 80.06 g/mol

Theo yield of SO₃ = (7.49 g S)(1 mol S/32.06 g)(2 mol SO₃/2 mol S)(80.06 g SO₃/mol) = <em>18.7 g SO₃</em>

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EleoNora [17]

A light layer of vacuum grease is applied to the rim of the belljar. Water at room temperature is placed inside and the vacuum pump is then used to evacuate the vessel. When the air pressure is reduced to the vapour pressure of water at room temperature the water will begin to boil.

5 0
3 years ago
Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). Match Term Definition Remo
pshichka [43]

Answer:

The answers to the questions are given below.

Explanation:

According to Le Chatelier's principle, if an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.

A. Effective of removing ammonia, NH3.

N2(g) + 3H2(g) ⇌ 2NH3(g)

Removing NH3 from the reaction simply means we are left with more reactants and no product. Therefore, the reactant will react to produce the product. Hence, the equilibrium position will shift to the right.

2. Effect of removing H2

N2(g) + 3H2(g) ⇌ 2NH3(g)

Remoing H2 simply means we have more products and less reactant. Therefore, the product will be convert to reactant. Hence, the equilibrium position will shift to the left.

C. Effect of adding a catalyst.

N2(g) + 3H2(g) ⇌ 2NH3(g)

Catalyst does not affect the equilibrium position. It only creates an alternative path to arrive at the product within a short time. Hence, it has no effect.

7 0
4 years ago
When carbon-containing compounds are burned in a limited amount of air, some CO(g) as well as CO₂(g) is produced. A gaseous prod
solong [7]

Here, we are going to calculate the mass % of C in the mixture.

What is a Mixture?

A mixture is composed of one or more pure substances in varying composition. There are two types of mixtures: heterogeneous and homogeneous. Heterogeneous mixtures have visually distinguishable components, while homogeneous mixtures appear uniform throughout.

Given that,

The mass % of CO =35.0% =35.0 g in 100 g mixture

The mass % of CO2 = 65% =65 g in 100 g mixture

Therefore,

The mass of C from CO = 15.007 g C

Similarly,

The mass of C from CO2 = 17.738 g C

Thus, the total mass of C = 15.007 g+17.738 g =32.745 g

Therefore,

The mass % of C= 32.745% =32.7%

Thus, the mass % of C in the mixture is 32.7%

To learn more about carbon-containing compounds click on the link below:

brainly.com/question/13381262

#SPJ4

3 0
2 years ago
Describe what you should do to recover only the water from a sample of muddy water.<br>​
xz_007 [3.2K]

Answer:

By using a filter.

Explanation:

4 0
3 years ago
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The steps in every scientific investigation must_______________________.
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Answer:

B. begin with a hypothesis

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