Molar mass ( CuSO₄) = 159.609 g/mol
159.609 g ----------------- 6.02 x 10²³ molecules
? g ------------------ 3.36 x 10²³ molecules
mass = ( 3.36 x10²³) x 159.609 / 6.02 x 10²³
mass = 5.36 x 10²⁴ / 6.02 x 10²³
mass = 8.90 g
hope this helps!
Answer:
Mass = 55.52 g
Explanation:
Given data:
Number of atoms of Li = 4.81×10²⁴ atom
Number of grams = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For Li:
4.81×10²⁴ atom × 1 mol / 6.022 × 10²³ atom
8 moles
Mass in gram:
Mass = number of moles × molar mass
Mass = 8 mol × 6.94 g/mol
Mass = 55.52 g
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
To learn more about pH here
brainly.com/question/15289741
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Answer:
The answer to your question is V = 0.108 L or 108 ml
Explanation:
Data
Volume = ?
mass = 0.405 g
Temperature = 273°K
Pressure = 1 atm
Process
1.- Convert mass of Kr to moles
83.8 g of Kr -------------------- 1 mol
0.405 g ------------------- x
x = (0.405 x 1) / 83.8
x = 0.0048 moles
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for V
V = nRT / P
- Substitution
V = (0.0048)(0.082)(273) / 1
- Simplification
V = 0.108 / 1
- Result
V = 0.108 L
Answer:
it’s D. to reduce indigestion
Explanation:
