As a ball bounces from a floor, its acceleration off the floor between bounces is

Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?

liberstina [14]

Answer : The maximum concentration of silver ion is $5\times 10^{-12}m$

Solution : Given,

$K_{sp}$ for AgBr = $5\times 10^{-13}$

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

$NaBr(aq)\rightleftharpoons Na^++Br^-$

The concentration of NaBr solution is 0.1 m that means,

$[Na^+]=[Br^-]=0.1m$

The equilibrium reaction for AgBr is,

$AgBr\rightleftharpoons Ag^++Br^-$

At equilibrium s s

The expression for solubility product constant for AgBr is,

$K_{sp}=[Ag^+][Br^-]$

The concentration of $Ag^+$ = s

The concentration of $Br^-$ = 0.1 + s

Now put all the given values in $K_{sp}$ expression, we get

$5\times 10^{-13}=(s)(0.1+s)$

By rearranging the terms, we get the value of 's'

$s=5\times 10^{-12}m$

Therefore, the maximum concentration of silver ion is $5\times 10^{-12}m$ .

4 0

2 years ago

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X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

3 years ago

True/False: If an object stays still, but the observer moves, the Doppler Effect is still observered

Vladimir79 [104]

Well, Harry, what you said is not necessarily true the way you said it.
But we know what you mean, and what you meant to say is true.

The Doppler effect is observed if there is relative radial motion
between an object and an observer <em><u>AND</u></em> if the object happens
to be putting out sound or light in the observer's direction.

3 0

3 years ago

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What is the length of the orbit of the earth around the sun?

Rzqust [24]

Answer:

S = 2 π R

R (mean) = 92.9E6 miles

S = 2 * 3.14 * 92.9E6 miles = 5.84E8 miles

5 0

2 years ago

6) what is ararge of speed

irina [24]

Answer:

in everyday use and in kinematic the speed of an object is the magnitude of the change of its position it is thus a scalar quantity

5 0

2 years ago