Answer: 3.59
Explanation:
(2.06)(1.743)(1.00)
2.06 × 1.743 × 1.00
= 3.59058
Two of the multiplied digits are represented in 3 significant figures. Therefore, for correct representation, the result of the product should be written to three significant figures.
3.59058 to 3 significant figures:
First three digits = 3.59
Fourth digit '0' is less than 5, and thus rounded to 0 with other succeeding digits
Therefore, (2.06)(1.743)(1.00) to 3 significant figures equals :
3.59
Answer:
We can make 10 percent solution by volume or by mass. A 10% of NaCl solution by mass has ten grams of sodium chloride dissolved in 100 ml of solution. Weigh 10g of sodium chloride. Pour it into a graduated cylinder or volumetric flask containing about 80ml of water.
Explanation:
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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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Answer:
Explanation:
NH3 + H3O+ --> NH4+ + H2O
equllibrium constant =K = [ H2O] [NH4+] / [NH3] [H3O+ ]
=
by inserting thier respecive values can you calcaulte, by the way coniseder [ H2O] =1 ,