All of the following require breaking and forming chemical bonds except

Arturiano [62]

Answer:

go to hellfhshahahahahahahahahaha aha

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7 0

3 years ago

A supersaturated solution can be prepared by dissolving solute in solvent while adding ______. A supersaturated solution contain

Delicious77 [7]

Answer:

Heat, solutes and high temperature.

Explanation:

A supersaturated solution can be formed by dissolving solute more solute in solvent by increasing temperature of the solution. A supersaturated solution contains more quantity of solutes than can be dissolved in the solvent at room temperature. A solution may remain supersaturated until the solution has high temperature and when the temperature started lower, the extra dissolve solutes begin undissolved and remain suspended in the solution.

4 0

3 years ago

What is the oxidizing agent in the following reaction? Cr2O3(2) + Al(s) > Cr(s) + Al2O(s)

zhenek [66]

$Al$ is a reducing agent, $Cr _2O _3$ is an oxidizing agent.

<h3>What is an oxidizing agent?</h3>

An oxidizing agent is a substance that causes oxidation by accepting electrons.

Al is a reducing agent, $Cr _2O _3$ is an oxidizing agent.

Hence, option C is correct.

Learn more about the oxidizing agent here:

brainly.com/question/10547418

#SPJ1

7 0

2 years ago

Metallic substances will conduct electricity in A) the liquid phase, only B) neither the solid nor the liquid phase C) the solid

Alenkinab [10]

Answer:

D) both the solid and the liquid phase

Explanation:

Metallic substances will conduct electricity in both the solid and liquid phase. Metals are generally conductors of heat and electricity.

In metals there are free mobile electrons the enables the flow electricity.
The free mobile electrons is utilized by the metals to conduct electricity.
Metals generally have metallic bonds that holds their particles together.
The metallic bonds enables metals to have a sea of delocalized electrons.
Metals have a large atomic radius and larger electron pool.

5 0

2 years ago

Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha

OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

$KH_2PO_4$ acts as a weak acid and $Na_2HPO_4$ as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

$H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2$

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

$7.00=6.86+log(\frac{base}{0.10})$