The Diagram Shows The Box For An Element In The Periodic Table What Is The Atomic Mass Of The Element Shown?

What is true about the formation of a bond?

timama [110]

Answer:

There is an overall release of energy when bonds form.

Explanation:

There is a general release of energy when bonds form. This energy is called bond energy.

Bond energy is involved in the breakdown or formation of one or more bonds between atoms of a molecule. Atoms bond with each other to achieve their electronic stability, that is, they move from a higher energy situation to a lower energy one. With this we can state that when the bond between atoms is formed, energy is released; therefore, its breakdown depends on energy absorption.

3 0

3 years ago

Determine the total number of atoms contain in a 2.00 moles of Ni

spayn [35]

Answer:

2.00 moles of Ni has 1.2 *10^24 atoms

Explanation:

Step 1: Data given

Number of moles Ni = 2.00 moles

Number of Avogadro = 6.022*10^23 /mol

Step 2: Calculate number of atoms

Number of particles (=atoms) = Number of Avogadro * number of moles

Number of atoms = 6.022 * 10^23 /mol * 2.00 moles

Number of atoms = 1.2*10^24 atoms

2.00 moles of Ni has 1.2 *10^24 atoms

4 0

3 years ago

2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na2C2O¬4, is the sodium salt of this acid.

AnnZ [28]

Answer:

Sodium oxalate is a basic salt. In water it can be dissolved and dissociated.

The oxalic acid in water has two dissociations.

Explanation:

Na2C2O4 ---> 2Na+ + C2O4-2

Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.

C2O4-2 + H2O ⇄ HC2O4- + OH-

In water the salt is basic. The pH of an aqueous solution of this salt is basic, since OH- is generated.

The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.

HC2O4- + H2O ⇄ H2C2O4 + OH-

The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).

H2C2O4 + H2O ⇄ H3O+ + HC2O4-

HC2O4- + H2O ⇄ H3O+ C2O4-2

The HC2O4- acts as an ampholyte since it accepts and delivers protons simultaneously.

6 0

3 years ago

Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.5

e-lub [12.9K]

Answer:

Buffer B has the highest buffer capacity.

Buffer C has the lowest buffer capacity.

Explanation:

An effective weak acid-conjugate base buffer should have pH equal to $pK_{a}$ of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.

Acetic acid is a weak acid and $CH_{3}COO^{-}$ is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):

$pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

$pK_{a}(CH_{3}COOH)=4.74$

Buffer A: $pH=4.74+log(\frac{0.10}{0.10})=4.74$

Buffer B: $pH=4.74+log(\frac{0.30}{0.30})=4.74$

Buffer C: $pH=4.74+log(\frac{0.10}{0.50})=4.04$

So, both buffer A and buffer B has same pH value which is also equal to $pK_{a}$ . Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.

The pH of buffer C is far away from $pK_{a}$ . Therefore, buffer C has the lowest buffer capacity.

6 0

3 years ago

A certain chemical reaction releases 31.0 kj/g of heat for each gram of reactant consumed. How can you calculate what mass of re

Aleksandr-060686 [28]

Answer:

See explanation.

Explanation:

Hello!

In this case, since we know the heat of reaction per gram of reactant and we should know the total energy of reaction, but it is not there, we are going to assume it is 1200 J as usual in these problems, so you can change it to whatever your given heat is.

In such a way, we set up the math as shown below:

$m= 1200J*\frac{1kJ}{1000J}*\frac{1g}{31.0kJ}$

Which results:

$m=0.0387g$

Best regards!

7 0

3 years ago