Answer:
umm it does not show what elements but I think its C. number of protons
2 HBr + Mg(OH) 2 yields Mg(Br) 2 + 2 H2O
M1V1 = M2V2
(0.245 M)(37.5 mL) = (M2)(18.0 mL)
9.1875 = (M2)(18.0 mL)
9.1875/(18.0 mL) = (M2)(18.0 mL)/(18.0 mL)
0.51041
this is not the complete answer as the mol to mol ratio must be considered!
For every 2 mols of HBr there is 1 mole of <span>Mg(OH)2; ratio = 2:1
</span>0.51041 / 2 = 0.25520
M2 = 0.25520
0.25520 M of<span> Mg(OH)2 solution</span>
Hope this helps and have a nice day!
Answer:
0.086
Explanation:
The formula for calculating the fraction of a covalent bond can be expressed as:
= exp (-0.25ΔE²)
![= exp[-0.25(E_{Al}-E_{p})^2]---(1)](https://tex.z-dn.net/?f=%3D%20exp%5B-0.25%28E_%7BAl%7D-E_%7Bp%7D%29%5E2%5D---%281%29)
from the equation above;
= the electronegativity of aluminum
= electronegativity of phosphorus
Using the data from periodic table figures;
= 1.5
= 2.1
∴
fraction of the covalent = exp[-0.25(1.5 - 2.1)²]
fraction of the covalent = exp(-0.09)
fraction of the covalent = 0.914
Now, the fraction of ionic bond will be = 1 - the fraction of covalent bond
= 1 - 0.914
∴
the fraction of bond that is ionic = 0.086