Answer: N = 2.78 × 10^23 atoms
There are N = 2.78 × 10^23 atoms in 70g of Au2cl6
Completed Question:
Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits
Explanation:
Given:
Molar mass of Au2cl6 = 303.33g/mol
Mass of Au2cl6 = 70g
Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol
According to the chemical formula of Au2cl6,
1 mole of Au2cl6 contains 2 moles of Au
Number of moles of Au = 2 × 0.231mol = 0.462mole
There are 6.022 × 10^23 atoms in 1 mole of an element.
Number of Atom of gold in 0.462 mole of gold is:
N = 0.462 mol × 6.022 × 10^23 atoms/mol
N = 2.78 × 10^23 atoms
Answer:
110 degree
Explanation:
This is because Hybridization of an s orbital with all three p orbitals (px , py, and pz) results in four sp3 hybrid orbitals. sp3 hybrid orbitals are oriented at bond angle of 109.5 degrees from each other. This 109.5 degrees gives an arrangement of tetrahedral geometry
Will be letter E cause all of them can combine molecule
Answer:
The volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.
Explanation:
For the complete recrystalization,
the amount of hot water should be such that, the benzoic acid is completely soluble in it.
As we are given that the solubility of benzoic acid in hot water is 68.0 g/L. Now we have to determine the volume of water is needed to recrystallize 0.700 g of benzoic acid.
we conclude that,
As, 68.0 grams of benzoic acid soluble in 1 L of water.
So, 0.700 grams of benzoic acid soluble in of water.
The volume of water needed = 0.01029 L = 10.29 mL
conversion used : (1 L = 1000 mL)
Therefore, the volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.
Answer:
50000ppm and 0.855M.
Explanation:
ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters
A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.
To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:
<em>mg NaCl:</em>
5g * (1000mg / 1g) = 5000mg
<em>L Solution:</em>
100mL * (1L / 1000mL) = 0.100L
ppm:
5000mg / 0.100L = 50000ppm
To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:
5g * (1mol / 58.5g) = 0.0855moles NaCl
Molarity:
0.0855mol NaCl / 0.100L = 0.855M