Answer:
0.0428 M
Explanation:
Because we're asked to calculate the molarity of nickel(II) cation, we need to <u>determine all sources for that species</u>, in this case, all Ni⁺² comes from the nickel(II) bromide solid (NiBr₂).
We use the molecular weight of NiBr₂ to calculate the moles of Ni:
1.87 g NiBr₂ ÷ 218.49g/mol * (1molNi⁺²/1molNiBr₂) = 8.55x10⁻³ mol Ni⁺²
Then we <u>divide the moles by the volume in order to calculate the concentration</u>:
8.55x10⁻³ mol Ni⁺² / 0.200 L = 0.0428 M
Asteroids are made up of metals and rocky material , while comets are made up of ice ,dust and rocky material.
The question is as follows: What is the% m / m of a solution in which 22 g of solute are dissolved in 44 g of solvent?
Answer: The% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.
Explanation:
Given: Mass of solute = 22 g
Mass of solvent = 44 g
The percentage m/m is calculated using the following formula.
Substitute the values into above formula as follows.
Thus, we can conclude that the% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.
9.5⋅<span>10<span>2
</span></span>Explanation:
To figure out how many atoms of copper you get in 1 gram of copper, you need to use copper's molar mass.
Answer:
<em>293.99 g </em>
OR
<em>0.293 Kg</em>
Explanation:
Given data:
Lattice energy of Potassium nitrate (KNO3) = -163.8 kcal/mol
Heat of hydration of KNO3 = -155.5 kcal/mol
Heat to absorb by KNO3 = 101kJ
To find:
Mass of KNO3 to dissolve in water = ?
Solution:
Heat of solution = Hydration energy - Lattice energy
= -155.5 -(-163.8)
= 8.3 kcal/mol
We already know,
1 kcal/mol = 4.184 kJ/mole
Therefore,
= 4.184 kJ/mol x 8.3 kcal/mol
= 34.73 kJ/mol
Now, 34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved in water.
For 101 kJ of heat would be
= 101/34.73
= 2.908 moles of KNO3
Molar mass of KNO3 = 101.1 g/mole
Mass of KNO3 = Molar mass x moles
= 101.1 g/mole x 2.908
= 293.99 g
= 0.293 kg
<em><u>293.99 g potassium nitrate has to dissolve in water to absorb 101 kJ of heat. </u></em>
