The `E_(M^(2+)//M)^(Theta)` for copper positive (+0.34V)It is only metal in the first series of transition elements showing this type of behevious. Explain.

The E^@(M^(2+)//M) value for copper is positive because:

The position of a particle is given by r=3.0r hati-2.0t^2 hatj+4.0 hatk m Where t is in seconds and the coefficients have the proper units for r to be in metres. Find the v and a of the particle?

Which transition metal of 3d series has positive E^@ (M^(2+)//M) value and why ?

A copper silver cell is set up. The copper ion concentration is 0.10M . The concentration of silver ion is not known. The cell potential when measured was 0.422V. Determine the concentration of silver ions in the cell. (Given) E_(Ag^(2+)//Ag)^@=+0.80V,E_(Cu^(2+)//Cu)^@=+0.34V

A partical of mass m1 is travelling with avelocity u1 strikes head-on against partical of mass m2 at rest.After collision the particales move with a velocity v1 and v2 respectively .show that the ratio of decrese in K.E of the partical of mass m1 to its original K.E is given By (4m1m2)/(m1+m2)^2

A famous relation in physics relates ‘moving mass' m to the ‘rest mass’ m_0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of s'pecial relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: m = m_0/(1-v^2)^(1/2)

The measured emf of the cell- Pt(s)H_2(G,1bar)|H^+(aq,1M)|| Cu^(2+)(aq,1M)|Cu(s)is 0.34V . What is the standard electrode potential of the half cell corresponding to the reaction?

One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknonw concentration. It other half cell consists of a zinc electrode dipping in 1.0M solution of Zn(NO_3)_2 . A voltage of 1.48V is measure for this cell. Use this information to calculate the concentration of silver nitrate solution used (E_(Zn^(2+)//Zn)^0=-0.76V,E_(Ag^+//Ag)^0=+0.80V)

The amplitude of current in series LCR circuit connected to an AC frequency "omega" is given by i_m = v_m/(sqrt(R^2+(X_L-X_C)^2) Where X_L and X_C are inductive and capacitative reactances respectively and "v_m" is amplitude of voltage. Starting fron this equation show that sharpness of resonance in the circuit is equal to the quality factor of the circuit.