Answer:
The Surface Of The Earth
Explanation:
The Troposphere contains about half of all the air in the entire atmosphere. Because it is at the bottom, air pressure, or the weight of the air, is greatest in this layer. (The surface of the earth is in this layer, making it the answer.)
The balanced equation for the above reaction is
2Al + 6H₂O ---> 2Al(OH)₃ + 3H₂
stoichiometry of Al to H₂ is 2:3
number of Al moles reacted - 78.33 g / 27 g/mol = 2.901 mol
according to molar ratio
2 mol of Al forms - 3 mol of H₂
therefore 2.901 mol of Al - forms 3/2 x 2.901 = 4.352 mol
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP
if 1 mol occupies 22.4 L
then 4.352 mol occupies - 22.4 L/mol x 4.352 mol = 97.48 L
volume occupied by H₂ is 97.48 L
Answer:
2.498g
Explanation:
First off it's important to know Avogadro's number relationship with number of moles and molar mass.
It is given as;
1 mole = 6.02 * 10^23 atoms.
A mole's weight is equal to it's molar mass. Hence the equation can be changed to;
1 mole = molar mass = 6.02 * 10^23 atoms
In this problem, molar mass of Sulphur is 32g/mol. This means 1 mole of sulphur weighs 32 g and contains 6.02 * 10^23 atoms. What mass of sulphur would then contain 4.7x10^22 of atoms?
This leads us to;
32 g = 6.02 * 10^23
x g = 4.7x10^22
Upon cross multiplication, we have;
x = (4.7x10^22 * 32) / 6.02 * 10^23
x = 24.98 * 10^-1
x = 2.498g
The liquid would evaporate
The molarity of a hydrochloric acid solution : 0.32 M
<h3>Further explanation </h3>
Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution).
Titrations can be distinguished including acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases is used.
Acid-base titration formula
Ma. Va. na = Mb. Vb. nb
Ma, Mb = acid base concentration
Va, Vb = acid base volume
na, nb = acid base valence
1 ⇒HCl (valence=1, HCl ⇒H⁺+Cl⁻, one H⁺)
2⇒Ca(OH)₂(valence=2, Ca(OH)₂⇒Ca²⁺+2OH⁻, two OH⁻)
M₂=0.1 M
V₂=48 ml=0.048 L
V₁=30 ml=0.03 L