Answer:
balanced in ACID not BASE
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Answer
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Explanation:
Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)
add H^1+ (acid) to capture the O and make 7 water molecules
Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O
Cr goes from +6 to +3 by gaining 3 e
Hg goes from 0 to +2 by losing 2 e
we need 3 Hg for every 2 Cr
so
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
2 Cr on the right and left
Net 12 positive charges on the right and the left
3 Hg on the right and left
14 H on the right and left
the equation is balanced
we cannot balance the equation in a basic solution with OH^1-
we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid
Answer:
10 kg of ice will require more energy than the released when 1 kg of water is frozen because the heat of phase transition increases as the mass increases.
Explanation:
Hello!
In this case, since the melting phase transition occurs when the solid goes to liquid and the freezing one when the liquid goes to solid, we can infer that melting is a process which requires energy to separate the molecules and freezing is a process that releases energy to gather the molecules.
Moreover, since the required energy to melt 1 g of ice is 334 J and the released energy when 1 g of water is frozen to ice is the same 334 J, if we want to melt 10 kg of ice, a higher amount of energy well be required in comparison to the released energy when 1 kg of water freezes, which is about 334000 J for the melting of those 10 kg of ice and only 334 J for the freezing of that 1 kg of water.
Best regards!
Answer:
1st order
Explanation:
because magnesium has a coefficient of 1, it is first order
hydrochloric acid has a coefficient of 2, so it is second order
overall the reaction is third order (1+2)
- See charge on ion is -1 .
Hence it has taken 1 electron
Now first look at EC of Fluorine(F)
- Now one electron added .hence no of electrons is 10now
Look at the EC
Or
![\\ \bull\sf\dashrightarrow [He]](https://tex.z-dn.net/?f=%5C%5C%20%5Cbull%5Csf%5Cdashrightarrow%20%5BHe%5D)
Option C is correct.
Ba 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² → [Xe]6s²
Ba - 2e⁻ → Ba⁺² [Xe]
