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strojnjashka [21]
2 years ago
11

Determine the molar solubility for Cr(OH)3 Ksp = 6.3x10^-31 ....

Chemistry
1 answer:
r-ruslan [8.4K]2 years ago
6 0

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What is a nuetron activation analysis
Daniel [21]
Its<span> a nuclear process used for determining the concentrations of elements in a cast amount of materials.</span>
8 0
3 years ago
What is the molarity of the solution resulting from the dissolution of 239 g glucose (C6H12O6) in 250
kifflom [539]

Answer:

Molarity =5.32 M

Explanation:

Given data:

Mass of glucose = 239 g

Volume = 250 mL (250 /1000 = 0.25 L)

Molarity = ?

Solution;

Formula:

Molarity = number of moles / volume in litter

Number of moles:

Number of moles = mass/ molar mass

Number of moles = 239 g / 180.2 g/mol

Number of moles = 1.33 mol

Molarity:

Molarity = number of moles / volume in litter

Molarity = 1.33 mol / 0.25 L

Molarity =5.32 M

6 0
3 years ago
compare the boiling point of water at sea leavel to the boiling point in denver (altitude = 1 mile) Explain
steposvetlana [31]

Answer:

boiling point decreases in denver

Explanation:

in higher places

theres less atmospheric pressure

it takes less energy to bring water to the boiling point.

Less energy means less heat

which means water will boil at a lower temperature

wonderopolisorg

6 0
1 year ago
What is the net ionic equation for a reaction between HCl and NaOH
kodGreya [7K]

Answer:

What is the net ionic equation for a reaction between HCl and NaOH?

Explanation:

A salt is a neutral ionic compound. Let's see how a neutralization reaction produces both water and a salt, using as an example the reaction between solutions of hydrochloric acid and sodium hydroxide. The overall equation for this reaction is: NaOH + HCl → H2O and NaCl

Hope that helped.

5 0
2 years ago
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i
omeli [17]

Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500 L

The concentration sodium sulfate = 2.104 M

Step 2: The equation

Na2SO4 → 2Na+ + SO4^2-

For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)

Step 3: Calculate the concentration of the ions

[Na+] = 2*2.104 M = 4.208 M

[SO4^2-] = 1*2.104 M = 2.104 M

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

8 0
2 years ago
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