→ ![2CrCl_3 + 3Co](https://tex.z-dn.net/?f=2CrCl_3%20%2B%203Co)
Explanation:
- The products formed are chromic chloride and cobalt.
Chromium + Cobaltous Chloride = Chromic Chloride + Cobalt
- Type of reaction is Single Displacement (Substitution) which is there is a displacement of one atom.
Reactants used in the reaction are -
- Chromium
![(Cr)](https://tex.z-dn.net/?f=%28Cr%29)
- Cobaltous Chloride
![(CoCl_2)](https://tex.z-dn.net/?f=%28CoCl_2%29)
Products formed in the reaction are -
- Chromic Chloride
![(CrCl_3)](https://tex.z-dn.net/?f=%28CrCl_3%29)
- Cobalt
![(Co)](https://tex.z-dn.net/?f=%28Co%29)
Hence, the chemical reaction is as follows -
→![CrCl_3 + Co](https://tex.z-dn.net/?f=CrCl_3%20%2B%20Co)
For balancing the above chemical equation we need to add a coefficient of 2 in front of chromium and of 3 in front of cobalt(II)chloride on right-hand-side while of 2 in front of chromium chloride and of 3 in front of carbon monoxide on left-hand-side of the equation.
Hence, the balanced equation is -
→ ![2CrCl_3 + 3Co](https://tex.z-dn.net/?f=2CrCl_3%20%2B%203Co)
Answer:
pH = 11.3
Explanation:
From the question given above, the following data were obtained:
Concentration of hydronium ion [H₃O⁺] = 4.950×10¯¹² M
pH =.?
The pH of a solution is defined by the following equation:
pH = –Log [H₃O⁺]
Thus, with the above formula, we can obtain the pH of the solution as follow:
Concentration of hydronium ion [H₃O⁺] = 4.950×10¯¹² M
pH =.?
pH = –Log [H₃O⁺]
pH = –Log 4.950×10¯¹²
pH = 11.3
Explanation:
high energy to Low energy
=the electron gains energy (K.E)
Answer:c
Explanation:
I think because ca^+2
It’s loses the ion and if u look back u would see that a cation is a t charge but it’s not Goan that electron it’s losing that electron
<u>Answer:</u> The energy of one photon of the given light is ![3.79\times 10^{-19}J](https://tex.z-dn.net/?f=3.79%5Ctimes%2010%5E%7B-19%7DJ)
<u>Explanation:</u>
To calculate the energy of one photon, we use Planck's equation, which is:
![E=\frac{hc}{\lambda}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
where,
= wavelength of light =
(Conversion factor:
)
h = Planck's constant = ![6.625\times 10^{-34}J.s](https://tex.z-dn.net/?f=6.625%5Ctimes%2010%5E%7B-34%7DJ.s)
c = speed of light = ![3\times 10^8m/s](https://tex.z-dn.net/?f=3%5Ctimes%2010%5E8m%2Fs)
Putting values in above equation, we get:
![E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{5.25\times 10^{-7}mm}\\\\E=3.79\times 10^{-19}J](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B6.625%5Ctimes%2010%5E%7B-34%7DJ.s%5Ctimes%203%5Ctimes%2010%5E8m%2Fs%7D%7B5.25%5Ctimes%2010%5E%7B-7%7Dmm%7D%5C%5C%5C%5CE%3D3.79%5Ctimes%2010%5E%7B-19%7DJ)
Hence, the energy of one photon of the given light is ![3.79\times 10^{-19}J](https://tex.z-dn.net/?f=3.79%5Ctimes%2010%5E%7B-19%7DJ)