Let the total mass of compound is 100g
The mass of each element will be
Al = 22.10 g
P = 25.40 g
O = 52.50 g
In order to determine the molecular formula we will calculate the molar ratio of the given elements
Atomic weight of Al : 27 g/ mol
Atomic weight of P : 3 1g /mol
Atomic weight of O : 16 g /mol
Moles of Al = mass / atomic mass = 22.10 / 27 = 0.819
Moles of P = mass / atomic mass = 25.40/ 31 = 0.819
Moles of O = mass / atomic mass = 52.50/ 16 = 3.28
Now we will divide the moles of each element with the lowest moles obtained to obtain a whole number ratio of moles of each element present
moles of Al = 0.819 / 0.819 = 1
moles of P = 0.819 / 0.819 = 1
moles of O = 3.28 / 0.819 = 4
So the empirical formula will be : AlPO4
The net ionic equation : Ba²⁺ + SO₄²⁻ ⇒ BaSO₄ (s)
<h3>Further explanation </h3>
The electrolyte in the solution produces ions.
The equation of a chemical reaction can be expressed in the equation of the ions
In the ion equation, there is a spectator ion that is the ion which does not react because it is present before and after the reaction
When these ions are removed, the ionic equation is called the net ionic equation
For gases and solids including water (H₂O) can be written as an ionized molecule
So only the dissolved compound is ionized ((expressed in symbol aq)
Barium sulfate can be formed from the reaction:
Ba(NO₃)₂(aq) + Na₂SO₄(aq)⇒BaSO₄(s)+2NaNO₃(aq)
For full ionic equation :
Ba²⁺ + 2NO₃⁻ + 2 Na⁺ + SO₄²⁻ ⇒ BaSO₄ (s) + 2 Na⁺ + 2 NO³⁻
by removing spectator ions (2NO₃⁻ and 2 Na⁺), the net ionic equation :
<em>Ba²⁺ + SO₄²⁻ ⇒ BaSO₄ (s) </em>
Answer:
The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s
Explanation:
Step 1: Data given
Molar mass helium = 4.0 g/mol
Molar mass O2 = 32 g/mol
Step 2: Graham's law
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of the molecular mass : 1/(Mr)^0.5
Rate of escape for He = 1/(4.0)^0.5 = 0.5
Rate of escape for O2 = 1/(32)^0.5 = 0.177
The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s
Step 1
<em>The reaction involved:</em>
CO + 2 H2 → CH3OH (completed and balanced)
------------
Step 2
<em>Data provided:</em>
19.7 g H2 (the limiting reactant)
Excess reactant = CO
144.5 g CH3OH = actual yield
----
<em>Data needed:</em>
The molar masses of:
H2) 2.00 g/mol
CH3OH) 32.0 g/mol
-----------
Step 3
The theoretical yield:
By stoichiometry,
CO + 2 H2 → CH3OH (The molar rate between H2 and CH3OH = 2:1)
2 x 2.00 g H2 --------- 32.0 g CH3OH
19.7 g H2 --------- X
X = 19.7 g H2 x 32.0 g CH3OH/2 x 2.00 g H2
X = 157.6 g CH3OH (The theoretical yield)
-----------
Step 4
The % yield is defined as follows:
Answer: d. 93% (it is the nearest value in comparison to my result)
