Answer:A mole is an arbitrary number of molecules in a single unit - refer to avogadro's number. Essentially, 1 mole is 6.022x10^23 molecules for ALL molecules or atoms, however one must remember that not all atoms/molecules are the same size, this is where mass comes into play. When you measure out 2 grams of carbon powder, there will be a lot more molecules present than if you weighed out 2 grams of thorium powder; this is because carbon is much smaller - kind of like a car filled with clowns, one given car can hold a lot of small clowns but only a few big ones; so the same volume is occupied but the amount of substance (clowns) varies on their own size. The arbitrary mass (relative to the hydrogen atom) for a molecule is the sum of its atomic components' atomic masses; e. g. C2H6's will have 2x12.00 (carbon) + 6x1.01 (hydrogen) = ~30 grams / mole.
Explanation:
Answer:
a
They form an ionic bond by exchanging two electrons.
Explanation:
Magnesium loses two electrons and oxygen gains these electrons so ionic bond is formed
Elements in the same group tend to have very similar properties (D). This is due to the number of valence electrons each group has.
Answer:
The mass of NaCl is 0.029 grams
Explanation:
Step 1: Data given
Molecular weight of NaCl = 58.44 g/mol
Volume of solution = 100 mL = 0.100 L
Molarity = 0.0050 M
Step 2: Calculate moles NaCl
Moles NaCl = molarity * volume
Moles NaCl = 0.0050 M * 0.100 L
Moles NaCl = 0.00050 moles
Step 3: Calculate mass NaCl
Mass NaCl = moles NaCl * molar mass NaCl
Mass NaCl = 0.00050 moles * 58.44 g/mol
Mass NaCl = 0.029 grams
The mass of NaCl is 0.029 grams
Answer:
You will have 19.9L of Cl2
Explanation:
We can solve this question using:
PV = nRT; V = nRT/P
<em>Where V is the volume of the gas</em>
<em>n the moles of Cl2</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is 273.15K assuming STP conditions</em>
<em>P is 1atm at STP</em>
The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:
63g * (1mol / 70.906g) = 0.8885 moles
Replacing:
V = 0.8885mol*0.082atmL/molK*273.15K/1atm
V = You will have 19.9L of Cl2
