Answer:
Explanation:
The atomic radius of elements are used to estimate the sizes of elements. The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.
To solve this problem we will obtain the atomic radius values of the given elements from a standard atomic radius table;
Si 111 pm
P 98 pm
Cl 79 pm
S 87pm
pm = picometer
We see that chlorine has the least atomic radius
Answer: 1.56 ATM
Explanation: if we assume temperature is constant, gas obeys
Boyles law pV= constant. Then p1·V1= p2·V2. And V1 = p2V2/p1
= 3.0 atm·0,52 l / 1.0 atm
Answer:
Atoms must have similar electronegativities in order to share electrons in a covalent bond.
Explanation:
Covalent bonding is one of the bondings that occurs between the atoms of elements. It is the bonding in which atoms share their valence electrons with one another. However, the ELECTRONEGATIVITY, which is the ability of an atom to be attracted to electrons play a major role in the formation of covalent bonds.
When atoms of different electronegativities combine, the more electronegative atom pulls more electrons towards itself, hence, an IONIC bond is formed. However, when the electronegativities of the atoms are similar, the sharing of their electrons becomes stronger. Hence, ATOMS MUST HAVE SIMILAR ELECTRONEGATIVITIES in order to share electrons in a covalent bond.
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
