1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anna11 [10]
3 years ago
13

Draw an energy diagram for an endothermic and exothermic reaction and label the diagram ​

Chemistry
2 answers:
just olya [345]3 years ago
5 0
Energy diagram for and endothermic and exothermic reaction

Molodets [167]3 years ago
5 0
Here they are.
Endothermic and exothermic reaction diagrams (labeled):

You might be interested in
In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].
Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose +  H^+\rightarrow  fructose+ glucose

The rate law of the reaction is given as:

R=k[H^+][sucrose]

[H^+]=0.01M

[sucrose]= 1.0 M

R=k[0.01M][1.0 M]..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

R'=[0.01 M][2.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

R'=[0.01 M][0.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when [H^+] is changed to 0.001 M = R'

R'=[0.0001 M][1.0 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}

R'=0.01\times R

If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and[H^+] both are changed to 0.1 M = R'

R'=[0.1M][0.1M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}

R'=1\times R

If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0
3 years ago
A runner completes a 5-mile run.how many yards did she run
slega [8]

The runner ran a total of 8800 yards.

5 0
3 years ago
Oh no... not again... Prof. Vitarelli spots Sybil running down the hall... yelling something... something about her tea cups...
Y_Kistochka [10]

Answer:

The reducing agent is Zn.

Explanation:

Let's consider the reaction between zinc and hydrochloric acid.

Zn(s) + 2 HCl(aq) ⇄ ZnCl₂(aq) + H₂(g)

This is a redox reaction, which can be divided in 2 half-reactions: reduction and oxidation.

In the reduction, H⁺ gains electrons and it is considered the oxidizing agent.

2H⁺ + 2 e⁻ ⇒ H₂

In the oxidation, Zn loses electrons and it is considered the reducing agent.

Zn ⇒ Zn²⁺ + 2 e⁻

6 0
3 years ago
The state of matter of the interior of the sun and other stars​
Sphinxa [80]

Answer: <u>It is, as all stars are, a hot ball of gas made up mostly of Hydrogen. The Sun is so hot that most of the gas is actually plasma, the fourth state of matter. ... As we heat up liquid, the liquid turns to gas. Gas is the third state of matter</u>

Explanation:

7 0
3 years ago
Read 2 more answers
A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was dil
Masteriza [31]

Answer : The value of pK_a of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}

\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol

Now we have to calculate the value of pK_a of the weak acid.

The equilibrium chemical reaction is:

                          HA+KOH\rightleftharpoons HK+H_2O

Initial moles     0.25     0.03        0

At eqm.    (0.25-0.03)   0.03      0.03

                     = 0.22

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[HK]}{[HA]}

Now put all the given values in this expression, we get:

3.85=pK_a+\log (\frac{0.03}{0.22})

pK_a=4.72

Therefore, the value of pK_a of the weak acid is, 4.72

7 0
3 years ago
Other questions:
  • Which of the following isotopes has the same number of neutrons as phosphorus – 21 (atomic number 15)
    10·1 answer
  • Which of the following is a reasonable ground state electron configuration
    12·1 answer
  • White blood cells, part of the ___________ system, work closely with the ___________ system to protect us from infection and dis
    5·1 answer
  • Name two metalloids that are semiconductors
    6·1 answer
  • The metal lithium has a diagonal relationship with the metal magnesium. true or false.
    9·1 answer
  • Why aluminium foils are use to wrap food items
    12·2 answers
  • The chemical bonding in sodium phosphate, Na3PO4, is classified as:
    14·2 answers
  • In this reaction, how does the rate of forward reaction vary with the concentration of the product? 2H2S(g) ⇌ 2H2(g) + S2(g) It
    12·2 answers
  • If a bowing ball hits a wall with a force of 6 N, the wall exerts a force of ____ back on the bowing ball
    6·1 answer
  • The unit of force newton is a derived unit.Why?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!