The sum of the masses of the reactants is equal to the sum of the masses of the products.
Answer:
V = 15.9512 dm³
Explanation:
Given data:
Pressure = P = 1.37 atm
Temperature = T= 315 K
Number of moles of nitrogen= n = 0.845 mol
Volume = V = ?
Formula:
PV = nRT
Now we will put the values in equation.
V = nRT/ P
V = ( 0.845 mol× 0.0821 dm³.atm.K⁻¹.mol⁻¹ × 315 K) / 1.37 atm
V = 21.853 dm³. atm/ 1.37 atm
V = 15.9512 dm³
ANSWER - 48.83%
Mg - 24.31
S - 32.07
O - 16x4 - 64
ADD ALL TOGETHER = 120.38
H - 1.01x2 - 2.02
O - 16
TOTAL = 18.02x7 = 126.14
ADD 120.38 to 126.14 = 246.52
last part divide MgSO4 by total compound mass
120.38/246.52 then multiply by 100 = 48.831
Answer:
The molarity of this solution is 14.82 mol/dm3 or 14.82 mol/L
Explanation:
- Molarity is the number of mole present in 1 Litre of solution. Molarity of a solution is a term referred to as concentration of a solution. The unit of Molarity is Mol/dm3 or Mol/L.
- let us make an assumption that the volume of the ammonia solution is 1L or 1dm3. Also, 1L = 1000 mL.
Step 1: calculate the mass of the solution
Density = 0.90g/ml (from the question)
Density = mass/ volume
Therefore Mass = density x volume
= 0.90g/ml x 1000ml
mass = 900 g
Step 2: calculate the mass of NH3 present in the solution
Since the concentrated aqueous of ammonia is 28%, It signifies that 1000ml of the solution contains 28% Ammonia
Recall from the above calculation that the mass of 1000 ml of solution is 900 g.
Therefore the mass of ammonia will be 28% of 900 g
mass of NH3 = 0.28 x 900 g
= 252 g
Step 3: calculate the number of mole of NH3
mole = mass/ molar mass
molar mass of NH3 = 17 g/mol
Therefore mole of NH3 = 252/17
= 14.82 mol
Step 4: Calculate Molarity
Molarity = number of moles/ volume of solution in Litre (L)
Molarity = 14.82 / 1
Molarity = 14.82 mol/L
The product of electrolysis : Al at cathode and Cl₂(chlorine) at anode
<h3>Further explanation</h3>
Given
Aluminum chloride compound
Required
The product of electrolysis
Solution
The rule :
The reaction at the cathode(the negative pole) :
1. the reduced active metal is water, other than that the metal will be reduced
2. H⁺ of the acid will be reduced
The reaction at the anode((the positive pole) :
1. if the electrodes are not inert then the metal is oxidized
2. If inert then:
a. OH⁻ from the base will be oxidized
b. The halogen metal will oxidize
In the electrolysis of molten AlCl₃ with an inert electrode, the cation will be reduced at the cathode and the anion will be oxidized at the anode
Cathode : Al³⁺ + 3e⁻ ⇒ Al(s)
Anode : 2Cl⁻⇒Cl₂(g) + 2e⁻
