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Triss [41]
3 years ago
8

Why is water classified as a renewable resource?

Chemistry
2 answers:
mrs_skeptik [129]3 years ago
6 0
The answer is a because the water we use it comes back again
777dan777 [17]3 years ago
6 0

Answer:

Option (A)

Explanation:

The resources that can be regenerated again over a specific period of time are known as renewable resources. For example, Wind, Water, and Sunlight.

Water is one of the most necessary and valuable renewable resources. It is considered to be renewable because it can be obtained again and again by undergoing certain processes that form a cycle which is commonly known as the water cycle.

In the water cycle, the water moves from the surface to the atmosphere in the form of vapor by undergoing the process of evaporation and transpiration and comes back to the surface in the form of rainfall, by undergoing the process of condensation.

This is how water changes its phase from one type to another, recycling within the system, thereby regenerating it. So, water is usually classified as a renewable resource.

Thus, the correct answer is option (A).

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According to the arrhenius theory an acid is a substance that
Vsevolod [243]

Answer:

An acid is a substance that releases H⁺ in aqueous solution.

Explanation:

There are different acid-base theories, such as Arrhenius, Bronsted-Lowry, Lewis, etc.

According to the Arrhenius theory, an acid is a substance that releases H⁺ in aqueous solution.

HCl(aq) → H⁺(aq) + Cl⁻(aq)

On the other hand, according to the Arrhenius theory, a base is a substance that releases OH⁻ in aqueous solution.

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

5 0
3 years ago
Read 2 more answers
Predict the products when aqueous solutions of sodium carbonate and calcium chloride are combined.
ikadub [295]
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6 0
2 years ago
Construct a three-step synthesis of 1,2-epoxycyclopentane from cyclopentanol by dragging the appropriate formulas into the bins.
zubka84 [21]

Answer:

(1) Bromination, (2) E2 elimination and (3) epoxidation

Explanation:

  • In the first step, -OH group in cyclopentanol is replaced by more facile leaving group Br by treating cyclopentanol with PBr_{3}
  • In the second step, E2 elimination in presence of strong base e.g. NaOEt/EtOH produce cyclopentene
  • In the third step, treatment of cyclopentene with mCPBA produces 1,2-epoxycyclopentane
  • Full reaction scheme has been shown below

3 0
2 years ago
Which analogy can best be likened to the activation energy of a chemical reaction? a drain through which water flows a slide dow
Nina [5.8K]
Answer is: <span>a hill over which a wagon is pushed.
</span>For all chemical reaction some energy is required and that energy is called activation energy (<span>energy that needs to be absorbed for a chemical reaction to start)<span>.
There are two types of reaction: endothermic reaction (chemical reaction that absorbs more energy than it releases) and exothermic reaction (chemical reaction that releases more energy than it absorbs).
</span></span>R<span>eactions occur faster with a catalyst because they require less activation energy.</span>
7 0
3 years ago
Read 2 more answers
Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d
worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.  

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B\rightarrow Products

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

3.20\times 10^{-1}=k(1.50)^a(1.50)^b ....(1)

Expression for rate law for second observation:

3.20\times 10^{-1}=k(1.50)^a(2.50)^b ....(2)

Expression for rate law for third observation:

6.40\times 10^{-1}=k(3.00)^a(1.50)^b ....(3)

Dividing 1 from 2, we get:

\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0

Dividing 1 from 3, we get:

\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[A]^1[B]^0

\text{Rate}=k[A]

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0
3 years ago
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