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
Alenkasestr [34]
4 years ago
8

Which of these statements explains the difference between nuclear binding energy and the strong nuclear force? Check all that ap

ply.
Nuclear binding energy holds the nuclear particles together

Nuclear binding energy is the energy needed to separate nuclear particles.

The strong nuclear force holds an atom’s protons and neutrons together.

The strong nuclear force is the energy released during radioactive decay processes.

The strong nuclear force is directly responsible for radioactive decay.

Nuclear binding energy can be calculated using E = mc2.
Chemistry
2 answers:
vitfil [10]4 years ago
8 0

Answer:Nuclear binding energy is the energy needed to separate nuclear particles

The strong nuclear force holds an atom’s protons and neutrons together

Nuclear binding energy can be calculated using E=mc2

Explanation:

EleoNora [17]4 years ago
6 0

Answer:

B, C, F.

Explanation:

I just took the test on Edge2020

You might be interested in
32,500 joules are used to boil a sample of water at its melting point what is the mass of the water?
Ivenika [448]

Answer:

43.18 gm

Explanation:

Heat of fusion of water at 0° C is 334  j/gm

 then you have to heat the water to boiling point (100 C)

    specific heat of water is 4.186 j/g-c

x = grams of water

x ( 334 + 4.186 (100) ) = 32500

x = 43.18 gm

3 0
2 years ago
A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn
nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of Ni^{2+} has fallen to 0.500 M is, 0.52 V

(c) The concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

E^0_{[Ni^{2+}/Ni]}=-0.23V

E^0_{[Zn^{2+}/Zn]}=-0.76V

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : Zn\rightarrow Zn^{2+}+2e^-     E^0_{[Zn^{2+}/Zn]}=-0.76V

Reaction at cathode (reduction) : Ni^{2+}+2e^-\rightarrow Ni     E^0_{[Ni^{2+}/Ni]}=-0.23V

The balanced cell reaction will be,  

Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}

E^o=(-0.23V)-(-0.76V)=0.53V

(a) Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}

E_{cell}=0.49V

(b) Now we have to calculate the cell potential when the concentration of Ni^{2+} has fallen to 0.500 M.

New concentration of Ni^{2+} = 1.50 - x = 0.500

x = 1 M

New concentration of Zn^{2+} = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}

E_{cell}=0.52V

(c) Now we have to calculate the concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}

Now put all the given values in the above equation, we get:

0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}

x=1.49M

The concentration of Ni^{2+} = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of Zn^{2+} = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0
4 years ago
A chemist wade out 5.96 grams of barium calculate the number of moles of barium she weighed out
PilotLPTM [1.2K]

Answer:

0.04 mol

Explanation:

Given data:

Mass of barium = 5.96 g

Moles of barium = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of barium = 5.96 g/ 137.33 g/mol

Number of moles = 0.04 mol

Thus the number of moles of barium in 5.96 g are 0.04 moles. The chemist weight out the 0.04 moles .

7 0
4 years ago
Which type of wave do the particles in a medium move parallel to the direction that the waves move
Talja [164]
Longitudinal waves. In a longitudinal wave the particles in the medium move parallel  to the direction waves go. A good example can be the p-waves in an earthquake.
8 0
4 years ago
2. Calculate the heat energy needed to raise the temperature of 30 grams of
Margaret [11]

Answer:

Really sorry I don't know answer

5 0
3 years ago
Other questions:
  • One mole of H atom is equal to
    5·1 answer
  • 61.0 mol of P4O10 contains how many moles of P
    5·2 answers
  • Why is there little heat retained in the warm thermosphere? a. The pressure is too low c. The molecules move too fast b. The pre
    8·2 answers
  • Hydrogen and Methanol have both been proposed as alternatives to hydrocarbon fuels. Write balanced reactions for the complete co
    5·1 answer
  • What is the boiling point of a solution made by mixing 75.0g ZnCl2 with 375.0 grams of water? (Kb for water is 0.512 C/m)
    12·1 answer
  • What is the conjugate base in the following equation: HBr + H2 O \longrightarrow ⟶ H3 O+ + Br-
    7·1 answer
  • I am doing an experiment that calls for 200g of H2SO4. How many moles of H2SO4 is this?
    10·1 answer
  • What is the bond polarity of SO3
    14·1 answer
  • WILL GIVE BRAINLIEST!:)
    5·1 answer
  • A chemist needs to find the concentration of a solution of barium hydroxide.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!