This is most likely "-100 kj/mol, spontaneous," but the question is difficult to read
Answer:
Part A:
Order: K<Rb<Cs
Part B:
Order: O<C<Be
Part C:
Order:CL<S<K
Part D:
Order:Mg<Ca<K
Explanation:
Atomic Size:
It is the distance from the center to atom to the valance shell electron. It is very difficult to measure the atomic size because there is definite boundary of atom.
Trend:
Moving from top to bottom in a group, Atomic Size increases.
Moving from left to right in a period, Atomic size generally decreases.
On the basis of above trend we will solve our question:
Part A:
All elements belong to Group 1:
Moving from top to bottom in a group, Atomic Size increases.
Order: K<Rb<Cs
Part B:
All elements belong to 2nd Period:
Moving from left to right in a period, Atomic size generally decreases.
Order: O<C<Be
Part C:
S belongs to 3rd Period, K, Cl belong to 4th period
Order:CL<S<K
Part D:
Mg is above Ca in group 2 and K is before Ca in 4th period
From trends described above:
Order:Mg<Ca<K
<h3>
Answer:</h3>
16.30 %
<h3>
Explanation:</h3>
- The percent by mass is calculated by dividing the mass of solute by the mass of solution and then multiplying by 100.
- That is. Percent by mass = (Mass of solute/Mass of solution) × 100
In this case;
Mass of solute = 14.60 g
Mass of solvent = 74.95 g
But, Mass of solution = mass of solute + mass of solvent
Thus;
Mass of solution = 14.60 g + 74.95 g
= 89.55 g
Therefore;
% mass = (14.60 ÷ 89.55 g) × 100%
= 16.30 %
Hence, the mass percent is 16.30%
Answer:
12.2 g
Explanation:
Step 1: Given data
Mass of O in KClO₄: 5.62 g
Step 2: Calculate the moles corresponding to 5.62 g of O
The molar mass of O is 16.00 g/mol.
5.62 g × (1 mol/16.00 g) = 0.351 mol
Step 3: Calculate the moles of KClO₄ that contain 0.351 moles of O
The molar ratio of KClO₄ to O is 1:4. The moles of KClO₄ that contain 0.351 mol of O are 1/4 × 0.351 mol = 0.0878 mol
Step 4: Calculate the mass corresponding to 0.0878 moles of KClO₄
The molar mass of KClO₄ is 138.55 g/mol.
0.0878 mol × (138.55 g/mol) = 12.2 g
Answer:
Explanation:
Given parameters:
Mass of CuCl₂ = 10.5g
Mass of Aluminium = 12.49g
Unknown:
Limiting reactant = ?
Solution:
The limiting reactant is the reactant given in short supply. Such reactant determines and controls the extent of the reaction.
Once they are used up, the reaction ceases to progress.
To find such reactant, we need to have a balanced reaction equation and the number of moles of the reactants:
3CuCl₂ + 2Al → 3Cu + 2AlCl₃
number of moles =
Molar mass of CuCl₂ = 63.5 + 2(35.5) = 134.5g/mol
Molar mass of Al = 27
Number of moles of CuCl₂ = = 0.08moles
Number of moles of Al = = 0.46moles
From the balanced reaction equation;
3 mole of CuCl₂ combined with 2 moles of Al
0.08 mole of CuCl₂ will combine with = 0.05moles of Al
But the given amount of Al is 0.46moles which is in excess.
Therefore CuCl₂ is in the limiting reactant.