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
Brrunno [24]
2 years ago
12

How can one differentiate between potential and potential difference????​

Chemistry
1 answer:
Eva8 [605]2 years ago
6 0

Electric Potential is the work done per unit charge in order to bring the charge from infinity to a point in electric field while Electric potential difference is the Potential developed while moving a charge from one point to another in the field itself.

You might be interested in
1. which is not a component of biodiversity
Vera_Pavlovna [14]
The answer is C, i just took the k-12 test~
8 0
3 years ago
Name the functional group in the following molecule:<br><br> H-C(Triple Bond)C-CH2CH3
Dima020 [189]

Here is your answer

Since, carbon atoms are triple bonded.

So functional group present in the compound is ALKYNE

HOPE IT IS USEFUL

3 0
3 years ago
In one sample of a compound of copper and oxygen, 3.12g of the compound contains 2.50g of copper and the remainder is oxygen. Ca
blsea [12.9K]

Answer:

% composition O = 19.9%

% composition Cu = 80.1%

Explanation:

Given data:

Total mass of compound = 3.12 g

Mass of copper = 2.50 g

Mass of oxygen = 3.12 - 2.50 = 0.62 g

% composition = ?

Solution:

Formula:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition Cu = (2.50 g / 3.12 g)×100

% composition Cu = 0.80 ×100

% composition Cu = 80.1%

For oxygen:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition O = (0.62 g / 3.12 g)×100

% composition O = 0.199 ×100

% composition O = 19.9%

5 0
3 years ago
rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is
Sunny_sXe [5.5K]

Complete Question

The  rate of a certain reaction is given by the following rate law:

            rate =  k [H_2][I_2]

rate Use this information to answer the questions below.

What is the reaction order in H_2?

What is the reaction order in I_2?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in H_2 is  n =  1

The reaction order in I_2 is  m = 1

The  overall reaction order z =  2

When the hydrogen is double the the initial rate is   rate_n  =  4.0*10^{-4} M/s

The rate constant is   k = 35.23 \  M^{-1} s^{-1}

Explanation:

From the question we are told that

   The rate law is  rate =  k [H_2][I_2]

   The rate of reaction is rate =  2.0 *10^{4} M /s

Let the reaction order for H_2 be  n and for I_2  be  m

From the given rate law the concentration of H_2 is raised to the power of 1 and this is same with I_2 so their reaction order is  n=m=1

   The overall reaction order is  

               z  = n +m

               z  =1 +1

               z  =2

At  rate =  2.0 *10^{4} M /s

        2.0*10^{4}  = k  [H_2] [I_2] ---(1)

= >    k  = \frac{2.0*10^{4}}{[H_2] [I_2]  }

given that the concentration of hydrogen is doubled we have that

            rate  = k [2H_2] [I_2] ----(2)

=>      k = \frac{rate_n  }{ [2H_2] [I_2]}

 So equating the two k

           \frac{2.0*10^{4}}{[H_2] [I_2]  } = \frac{rate_n  }{ [2H_2] [I_2]}

    =>    rate_n  =  4.0*10^{-4} M/s

So when

      rate_x =  52.0 M/s

        [H_2] = 1.8 M

         [I_2] =  0.82 \ M

We have

      52 .0 =  k(1.8)* (0.82)

     k = \frac{52 .0}{(1.8)* (0.82)}

      k = 35.23 M^{-2} s^{-1}

     

     

3 0
3 years ago
Consider the single‑step, bimolecular reaction. CH3Br+NaOH⟶CH3OH+NaBr When the concentrations of CH3Br and NaOH are both 0.120 M
GarryVolchara [31]

Answer:

Explanation:

CH₃Br+NaOH⟶CH₃OH+NaBr

It is a single step bimolecular reaction so  order of reaction is 2 , one for CH₃Br and one for NaOH .

rate of reaction = k x [CH₃Br] [ NaOH]

.008 = k x .12 x .12

k = .55555

when concentration of CH₃Br is doubled

rate of reaction = .555555 x [.24] [ .12 ]

= .016 M/s

when concentration of NaOH  is halved

rate of reaction = .555555 x [.12] [ .06 ]

= .004  M/s

when concentration of both CH₃Br and Na OH is made 5 times

rate of reaction = .555555 x .6 x .6

= 0.2 M/s

8 0
3 years ago
Other questions:
  • Why are chemical tests necessary to tell a dextrose solution from a saline solution?
    14·1 answer
  • Which substance is the least likely to dissolve in water?
    15·2 answers
  • Covert from cm to m plz and thank u:);)
    15·2 answers
  • What particles are affecting by strong nuclear forces?
    13·1 answer
  • A car with a 14 gallon gas tank gets 22 miles per gallon. How far can the car travel on a full tank of gas?
    15·1 answer
  • Ocean life is changing due to<br>This is happening because<br>This is causing<br>because​
    12·1 answer
  • Why are diamonds unevenly distributed on Earth?
    8·1 answer
  • DATA SHEET p 45. TRIAL 1 TRIAL 2 1. Mass of the ground pretzel 1.00 gram 1.03 g 2. Initial volume of the AgNO3 solution 0.00 mL
    12·1 answer
  • 4. One gram of of ............... contains one type of atoms. a. iron b. ammonia c. hydrogen chloride d. water​
    13·1 answer
  • How many atoms are in 76.15 grams of fe<br><br> urgently needed, will mark brainliest
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!