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vladimir2022 [97]
3 years ago
9

1. Which of the following choices demonstrates the law of constant composition? (Slides 2 ‒ 3: Laws on Matter) (a) Nitrogen and

oxygen are both found in nature as diatomic molecules. (b) When 20.0 g of nitrogen and 32.0 g of oxygen are combined and allowed to react in two separate experiments, both times the product isolated from reaction contains 14.0 g of nitrogen and 32.0 g of oxygen. (c) Nitrogen and oxygen gases are mixed to produce a sample consisting of 20.0 g of nitrogen and 32.0 g of oxygen. (d) When 14.0 g of nitrogen and 32.0 g of oxygen are mixed and react, there is no measurable change in mass during the reaction. (e) Nitrogen and oxygen are gases found in air, nitrogen is approximately 79% of air and oxygen is approximately 21%.
Chemistry
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

(b) When 20.0 g of nitrogen and 32.0 g of oxygen are combined and allowed to react in two separate experiments, both times the product isolated from reaction contains 14.0 g of nitrogen and 32.0 g of oxygen.

Explanation:

The law of definite proportion states that a gen chemical compound always contains its constituent elements in a fixed ratio by mass, independent on the method of preparation.

The molar mass of Nitrogen and Oxygen would always remain the same, allowing for exact reactant masses (or mole ratio) irrespective of the given amount of sample.

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Explanation:

A chemical reaction is defined as the reaction in which bonds between the reactants either break or form which leads to the formation of a new substance.

For example, 2Li + Cl_{2} \rightarrow 2LiCl

So, when we drop a sodium metal into water then it produces a frizzing sound which shows the metal is reacting with water.

We know that when two aqueous solutions chemically react with each other then it may lead to the formation of an insoluble substance which is known as precipitate.

This means that formation of a precipitate is also a chemical reaction.

Thus, we can conclude that following are the statements which show evidence for a chemical reaction.

  • Dropping sodium metal into water produces fizzing.
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Compare and contrast the movement of electric charge in a solution with the transfer of electric charge between solid objects
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The idea of electric field was presented by Michael Faraday. The electrical field constrain acts between two charges, similarly that the gravitational field compel acts between two masses.
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4 years ago
If an object is accelerating at a rate of 25 m/s2, how long (in seconds) will it take to reach a speed of 550 m/s? (Assume an in
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22s

Explanation:

Given parameters:

Acceleration  = 25m/s²

Final velocity = 550m/s

Initial velocity  = 0m/s

Unknown:

Time taken  = ?

Solution:

To solve this problem, we are going to use one of the motion equations:

                 v  = u + at

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

                  550 = 0 + 25 x t

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8 0
3 years ago
What is an acid?
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7 0
3 years ago
Using 3 O2 molecules and 5 H2 molecules, how many water molecules can be produced? Do you have any left over?
makvit [3.9K]

Answer:

5 molecules of H₂O can be produced

0.5 molecules of O₂ did not reacted

Explanation:

The reaction is:  2H₂(g)  +  O₂(g)  →  2H₂O (g)

Firstly we determine the limiting reactant:

2 moles of hydrogen need 1 mol of oxygen to react

We must know the moles of each.

6.02ₓ10²³ molecules is 1 mol

3 molecules are ____ 3 /6.02ₓ10²³ = 4.98×10⁻²⁴ moles O₂

5 molecules are ____ 5 / 6.02ₓ10²³ = 8.30×10⁻²⁴ moles H₂

2 moles of H₂ need 1 mol of O₂

Then 8.30×10⁻²⁴ moles of H₂ must need (8.30×10⁻²⁴ .1) / 2 = 4.15×10⁻²⁴ moles O₂. It is ok, because I have 4.98×10⁻²⁴ moles O₂. Oxygen is the reagent in excess, so the limiting is the H₂

1 moles of O₂ needs 2 moles of H₂ to react

Then, 4.98×10⁻²⁴ moles of O₂ must need (4.98×10⁻²⁴ .2) / 1 =9.96×10⁻²⁴ moles of H₂, we don't have enough H₂

So, in the reaction ratio is 2:2.

8.30×10⁻²⁴ moles of H₂ will produce 8.30×10⁻²⁴ moles H₂O

1 mol has 6.02×10²³ molecules

8.30×10⁻²⁴ must have (8.30×10⁻²⁴ . NA) = 5 molecules

The reagent in excess is the O₂. These means that there is oxygen that has not reacted.

We have 4.98×10⁻²⁴ moles O₂ and we used 4.15×10⁻²⁴ moles.

(4.98×10⁻²⁴ - 4.15×10⁻²⁴) = 0.83×10⁻²⁴ moles of oxgen hasn't reacted.

1 mol is contained by NA molecules

0.83×10⁻²⁴ moles are contained by (0.83×10⁻²⁴ . 6.02×10²³) = 0.5 molecules

6 0
3 years ago
Read 2 more answers
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