Metallic solids or metallic structures experience metallic bonds which are the forces of attractions between the sea of electrons and the nucleus of the metallic atoms. They share a network of highly delocalized electrons.
I therefore think that the packing efficiency decreases as the number of nearest neighbors decreases.
It will have 35 ''electrons'' . Basically the number of protons in the nucleus of an atom is always equal to the number of electrons but its just that protons are positively charged and electrons are negatively charged. <span />
Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
Answer:
2
Explanation:
to separate objects or ideas into group based on ways they are alike
Answer:
The pH value of the mixture will be 7.00
Explanation:
Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,
![pH=pK_{a} + log(\frac{[Base]}{[Acid]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%29)
According to the given conditions, the equation will become as follow
![pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BNa_%7B2%7DHPO_%7B4%7D%20%5D%7D%7B%5BNaH_%7B2%7DPO_%7B4%7D%5D%7D%29)
The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.
Placing all the given data we obtain,
