According to the octet rule, atoms tend to gain, lose, or share electrons until they are surrounded by__8__ valence electrons.
First we have to convert:145 pm = 145 * 10^(-12) m36 cm = 360 mm = 360 * 10^(-3) mn = 360 * 10^(-3) m / 145 * 10^(-12) m = = 360 * 10^(-3) * 10^(12) / 145 = = 2.482758621 * 10^(9) or:2,482,758,621 atoms.
Answer:
8.0 moles
Explanation:
Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.
Using the formula: 
Concentration of acid = ?
Volume of acid = 10 mL
Concentration of base = 1.0 M
Volume of base = 40 mL
mole of acid = 1
mole of base = 1
Substitute into the equation:
Concentration of acid = 40/10 = 4.0 M
To determine the number of moles of acid present in 2.0 liters of the unknown solution:
Number of moles = Molarity x volume
molarity = 4.0 M
Volume = 2.0 Liters
Hence,
Number of moles = 4.0 x 2.0 = 8 moles
Answer:
2.94x10²² atoms of Cu
Explanation:
We must work with NA to solve this, where NA is the number of Avogadro, number of particles of 1 mol of anything.
Molar mass Cu = 63.55 g/mol
Mass / Molar mass = Mol → 3.11 g / 63.55 g/m = 0.0489 moles
1 mol of Cu has 6.02x10²³ atoms of Cu
0.0489 moles of Cu, will have (0.0489 .NA)/ 1 = 2.94x10²² atoms of Cu
The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
