Answer:
Explanation:
K₂CrO₄ + ( COONa )₂ + 2BaCl₂ = Ba CrO₄ + ( COO ) ₂ Ba + 2 KCl + 2 NaCl
.033 M .053 M
Ksp of Ba CrO₄ is 2.10×10⁻¹⁰
Ksp of ( COO ) ₂ Ba is 1.30×10⁻⁶
A ) Ksp of Ba CrO₄ is less so it will precipitate out first .
B) Ksp = 2.10×10⁻¹⁰
Ba CrO₄ = Ba⁺² + CrO₄⁻²
C .033
C x .033 = 2.10×10⁻¹⁰
C = 63.63 x 10⁻¹⁰ M
Ba⁺² must be present in concentration = 63.63 x 10⁻¹⁰ M
C)
90% of precipitation of barium oxalate
concentration of oxalate to precipitate out = .9 x .0532 = .04788
( COO ) ₂ Ba = (COO)₂⁻² + Ba⁺²
.04788 M C
C x .04788 = 1.30×10⁻⁶
C = 27.15 x 10⁻⁶ M .
In Silver, the 4d orbitals will be completely filled. That
implies that it does not have two electrons in the 5s orbital. The electronic
configuration of Silver is :
Ag (47) = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^1
4d^10
As another example, consider the downward force<span> of gravity that the Earth exerts on </span>you<span> (also </span>called<span> weight). In turn, </span>you<span> exert an exactly equal upward </span>force<span> on the Earth. Together, those </span>two forces<span> form an action-reaction </span>force pair<span>.</span>
The energy transferred is 28.5 j
<em><u>calculation</u></em>
energy is calculate MCΔT formula where,
M(mass)= 6.30 grams
C(specific heat capacity)= 0.377 j/g
ΔT(change in temperature)= 32.0c- 20 c= 12 c
Energy is therefore= 6.30 g x 0.377 j/g /c x 12 c =<u>28.5 j</u>
Explanation:
since it has been accelerated to 5℅ of the sleec of light then
v= 5/100 × (3× 10^8)
after getting velocity you'll substitute the values in the formula
de Broglie wavelength= h/mv
don't forget to change the grams in mass to kg
aorry I don't hav my calc with me
