Answer:
0.11mole
Explanation:
Let us assume that the condition is at standard temperature and pressure(STP);
Given parameters:
Volume of water = 2.45L
Unknown:
Number of moles found in this volume of water = ?
Solution;
At STP;
Number of moles = 
Input the parameters and solve;
Number of moles of water = 
= 0.11mole
The number of moles of water found is 0.11mole
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.
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Avogadros number is the answer
The mass of silver chloride produced is 24.79g.
Given,
volume = 1.11 L
Molarity = 0.156 M
First, convert molarity to moles by multiplying by the volume:
0.156 M AgNO3 = (0.156 moles AgNO3)/1 L x 1.11 L = 0.173 moles AgNO3
We are only interested in AgCl, not AgNO3. We convert moles of silver nitrate to moles of silver chloride using the balanced equation. This is a 1:1 conversion:
0.173 moles AgNO3 x (1 mole AgCl)/(1 mole AgNO3) = 0.173 moles AgCl
To get mass from moles, we multiply by the molar mass:
0.173 moles AgCl x (143.32 g/mol) = 24.79 g AgCl
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Answer:
The embryo, endosperm, and seed coat are the three major parts of a seed. An embryo is the young multicellular organism formed before it emerges from the seed. A seed is an embryonic plant, which stores food and is enclosed in a protective outer covering, which give rise to a new plant.
