Answer:
The ground state configuration is the lowest energy, most stable arrangement. An excited state configuration is a higher energy arrangement (it requires energy input to create an excited state). Valence electrons are the electrons utilised for bonding.
or the
FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.
Element Atomic number Electron configuration
sulfur 16 1s22s22p63s23p4
chlorine 17 1s22s22p63s23p5
argon 18 1s22s22p63s23p6
or the
Two electrons
Two electrons fill the 1s orbital, and the third electron then fills the 2s orbital. Its electron configuration is 1s22s1.
Explanation:
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In order to find the mass of tin with the given volume of 5.5 L and density of 7.265, we will use the formula
Density = Mass / Volume
We will just multiply both sides of the equation by the volume and we will get:
Mass = Volume x Density
We can now solve the problem by substituting the given.
Mass = 5.5 L x 7.265 g/L
Mass = 39.96 g
Answer: there are 39.96 grams of tin
Answer:
mass = 242.372 grams
Explanation:
1- getting the number of moles of HCl:
molarity = number of moles of solute / liters of solution
4 = number of moles of HCl / 5.2
number of moles of HCl = 4 * 5.2 = 20.8 moles
2- getting the number moles of magnesium:
From the balanced equation given, we can note that one mole of magnesium is required to react with two moles of HCl. To get the number of moles required to react with 20.8 moles of HCl. we will simply do cross multiplication as follows:
1 mole of Mg ...............> 2 moles of HCl
?? moles of Mg ...........> 20.8 moles of HCl
Number of moles of Mg = 20.8 / 2 = 10.4 moles
3- getting the mass of Mg:
number of moles = mass / molar mass
Using the periodic table, we can find that the molar mass of magnesium is 24.305 grams.
This means that:
10.4 = mass / 24.305
mass = 24.305 * 10.4
mass = 242.372 grams
Hope this helps :)
The silver ion concentration in saturated solution of silver (i) phosphate is calculated as follows.
write the equation for dissociation of silver (i) phosphate
that is Ag3PO4 (s) = 3Ag^+(aq) + PO4 ^3-(aq)
let the concentration of the ion be represented by x
ksp is therefore= (3x^3 )(x) = 1.8 x10 ^-18
27 x^3 (x) = 1.8 x10^-18
27x^4 = 1.8 x10^-18 divide both side 27
X^4 = 6.67 x10 ^-20
find the fourth root x = 1.6 x10 ^-5m
the concentration of silver ion is therefore = 3 x (1.6 x10^-5) = 4.8 x10^-5m
