What is chemical energy? (2 points)a
Energy that is stored or released during a chemical reaction

b
Energy that is destroyed during a physical change

c
Energy that is released during a phase change

d
Energy that is created when a substance is being burned

Answers

Answer 1
Answer:

Final answer:

Chemical energy is the energy that is stored or released during a chemical reaction. It exists within the chemical bonds of molecules and can be converted into other forms of energy.

Explanation:

Chemical energy refers to the energy that is stored or released during a chemical reaction. It is a form of potential energy that exists within the chemical bonds of molecules. When a chemical reaction takes place, these bonds are broken or formed, resulting in the release or absorption of energy.

For example, when a fuel such as gasoline is burned, the chemical energy stored in the hydrocarbon molecules is converted into heat and light energy. Similarly, during photosynthesis, plants convert sunlight into chemical energy in the form of glucose.

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Answer 2
Answer:

Explanation:

Chemical energy

Chemical energy is energy stored in the bonds of chemical compounds, like atoms and molecules. This energy is released when a chemical reaction takes place.

Usually, once chemical energy has been released from a substance, that substance is transformed into a completely new substance.


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HELP if water vapor condense on a cold surface is the initial and final energy thermal, phase, or chemical?

Answers

My answer to this great question would be thermal

Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. ResetHelp 1. NaCl ; ionic bonds{\rm NaCl} ; blank are stronger than the blank in {\rm HCl}. are stronger than the dispersion forces{\rm NaCl} ; blank are stronger than the blank in {\rm HCl} . in HCl. 2. H2O ; hydrogen bonds{\rm H_2O} ; blank are stronger than the blank in {\rm H_2Se}. are stronger than the dispersion forces{\rm H_2O} ; blank are stronger than the blank in {\rm H_2Se}. in H2Se. 3. NH3 ; hydrogen bonds{\rm NH_3} ; blank are stronger than the blank in {\rm PH_3}. are stronger than the dipole-dipole attractions{\rm NH_3} ; blank are stronger than the blank in {\rm PH_3}. in PH3. 4. HF ; hydrogen bonds{\rm HF} ; blank are stronger than the blank in {\rm F_2}. are stronger than the dispersion forces{\rm HF} ; blank are stronger than the blank in {\rm F_2}. in F2.

Answers

Answer:

The ionic bond in NaCl are stronger than the stronger than the dispersion forces in HCl.

The hydrogen bonds in H2O are stronger than the dispersion forces in H2Se

Hydrogen bonds in NH3 are stronger than the dipole-dipole attractions in PH3.

Hydrogen bonds in HF are stronger than the dispersion forces in F2

Explanation:

Ionic bonds occur in molecules with high differences in their electronegative value where there are actual transfer of electrons. HCl has a bond which is involved in the sharing of electrons.

Hydrogen bonds are present in H2O which is stronger than the dispersion forces.

PH3 is a larger molecule with greater dispersion forces than ammonia, NH3 has very polar N-H bonds leading to strong hydrogen bonding. This dominant intermolecular force results in a greater attraction between NH3 molecules than there is between PH3 molecules.

F2 is a non-polar molecule, therefore they have London dispersion forces between molecules while HF has a hydrogen bond because F is highly electronegative.

Final answer:

Ionic bonds are stronger than dispersion forces in HCl, while hydrogen bonds are stronger than dispersion forces in H2Se, PH3, and F2.

Explanation:

In the given sentences, the blanks represent the types of intermolecular forces. The options given are ionic bonds, hydrogen bonds, dispersion forces, and dipole-dipole attractions. Ionic bonds are stronger than the dispersion forces in HCl. Hydrogen bonds are stronger than the dispersion forces in H2Se. Hydrogen bonds are stronger than the dipole-dipole attractions in PH3. Hydrogen bonds are stronger than the dispersion forces in F2.

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At 445 °C, Kc = 50.2. If one starts with 0.100 M H2 (g), 0.100 M I2 (g) and 0.0500 M HI (g) what is the equilibrium concentration of HI?

Answers

Here's The Answer:  K = 50.2 = (2x)^2 / (0.1-x)^2 
x = 0.078
so H2 eq = 0.022 M

Hope this helped! :D

A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equivalence. The pKb of aniline is 4.87 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added.

Answers

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of HNO_3

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of C_6H_5NH_2.

M_2\text{ and }V_2 are the final molarity and volume of HNO_3.

We are given:

M_1=0.3403M\nV_1=160.0mL\nM_2=0.0501M\nV_2=?

Putting values in above equation, we get:

0.3403M* 160.0mL=0.0501M* V_2\n\nV_2=1086.79mL

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of C_6H_5NH_2 +  Volume of HNO_3

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

\text{Concentration of salt}=(0.3403M)/(1246.79mL)* 160.0mL=0.0437M

Now we have to calculate the pH of the solution.

At equivalence point,

pOH=(1)/(2)[pK_w+pK_b+\log C]

pOH=(1)/(2)[14+4.87+\log (0.0437)]

pOH=8.76

pH+pOH=14\n\npH=14-pOH\n\npH=14-8.76\n\npH=5.24

Thus, the pH of the solution is, 5.24

What is the most likely position for the hurricane indicated by the wind readings from the three weather stations shown?

Answers

Here's the answer, I remember doing this problem last year.

23.5 degrees north, 77 degrees west

If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

Answers

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide