Mg (s) ---> Mg²⁺ (aq) + 2e
Co²⁺ (aq) + 2e ---> Co(s)
Cu (s) ---> Cu²⁺ (aq) + 2e
Au²⁺ (aq) + 2e ---> Au(s)
The bond in NO3- ion is polar covalent and oxygen pulls the electrons.
A polar covalent bond is formed when there is a significant difference in electronegativity between covalently bonded atoms.
Such bonds are said to be polar covalent with the electron density tilted towards one of the bonding atoms.
In NO3-, Nitrogen and oxygen have an electronegativity difference of 0.5, so the bond is polar covalent, with oxygen pulling the electrons toward it.
Learn more;brainly.com/question/25150590
Answer : The enthalpy change is, 7.205 KJ
Solution :
The conversions involved in this process are :
![(1):H_2O(s)(-25^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(80^oC)](https://tex.z-dn.net/?f=%281%29%3AH_2O%28s%29%28-25%5EoC%29%5Crightarrow%20H_2O%28s%29%280%5EoC%29%5C%5C%5C%5C%282%29%3AH_2O%28s%29%280%5EoC%29%5Crightarrow%20H_2O%28l%29%280%5EoC%29%5C%5C%5C%5C%283%29%3AH_2O%28l%29%280%5EoC%29%5Crightarrow%20H_2O%28l%29%2880%5EoC%29)
Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of water = 10 g
= specific heat of solid water = 2.09 J/gk
= specific heat of liquid water = 4.18 J/gk
n = number of moles of water = ![\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{10g}{18g/mole}=\frac{10}{18}mole](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BMass%20of%20water%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20water%7D%7D%3D%5Cfrac%7B10g%7D%7B18g%2Fmole%7D%3D%5Cfrac%7B10%7D%7B18%7Dmole)
= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get
![\Delta H=[10g\times 2.09J/gK\times (273-248)k]+\frac{10}{18}mole\times 6010J/mole+[10g\times 4.18J/gK\times (353-273)k]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B10g%5Ctimes%202.09J%2FgK%5Ctimes%20%28273-248%29k%5D%2B%5Cfrac%7B10%7D%7B18%7Dmole%5Ctimes%206010J%2Fmole%2B%5B10g%5Ctimes%204.18J%2FgK%5Ctimes%20%28353-273%29k%5D)
(1 KJ = 1000 J)
Therefore, the enthalpy change is, 7.205 KJ
Answer: -
2 x 10 ²⁴ atoms of Ag rounded off to 1 significant figures.
Explanation: -
Mass of Cu = 100 g
Molar mass of Cu = 63.55 g / mol
Number of moles of Cu = 100 g / (63.55 g/mol)
= 1.574 mol of Cu
The balanced chemical equation for the reaction is
Cu + 2AgNO₃ → Cu(NO₃)₂ + 2 Ag
From the balanced chemical equation we see
1 mol of Cu gives 2 moles of Ag
1.574 mol of Cu gives ![\frac{2 moles of Ag x .574 mol of Cu}{1 mo of Cu}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%20moles%20of%20Ag%20x%20.574%20mol%20of%20Cu%7D%7B1%20mo%20of%20Cu%7D%20%20)
= 3.148 mol of Ag
1 mole of Ag has 6.02 x 10 ²³ atoms of Ag according to Avogadro's Law.
3.148 mol of Ag has ![\frac{6.02 x 10 ²³ atoms of Ag x 3.148 mol of Ag}{1 mol of Ag}](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.02%20x%2010%20%C2%B2%C2%B3%20atoms%20of%20Ag%20x%203.148%20mol%20of%20Ag%7D%7B1%20mol%20of%20Ag%7D%20%20)
= 18.95 x 10²³ atoms of Ag
= 2 x 10 ²⁴ atoms of Ag rounded off to 1 significant figures.