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3 years ago

What color is the 2,6-dichloroindophenol solution in acidic solutions?

1 answer:

7 0

DCPIP is a redox dye that starts off blue but will turn reddish-pink in acidic solution.

This chemical is often used in general chemistry experiments with vitamin C (ascorbic acid), which is a good reducing agent. The DCPIP become colorless when it is reduced (remember, it's blue when oxidized).

You might be interested in

5. How many atoms of hydrogen are in a molecule of acetic acid (HC2H202)?

Kazeer [188]

Answer:

in acetic acid ,there are 4 hydrogen atoms

Explanation:

C H 3 C O O H

3 0

3 years ago

In the forward reaction of this equilibrium, which substance acts as the brønsted-lowry base? h2s(aq)+ch3nh2(aq)⇋hs−(aq)+ch3nh+3

spin [16.1K]

<span>Answer: Bronsted base is something that accepts proton (H+) and acid is something that donates H+ so here CH3NH2 will be the base and H2S is the acid.</span>

7 0

3 years ago

Sodium hydroxide is a common ingredient in drain cleaners such as Drano. The mole fraction of sodium hydroxide in a saturated aq

SVETLANKA909090 [29]

Answer:

The correct answer to this question is (c) 25.0 m

Explanation:

To solve this we list out the variables thus

mole fraction of sodium hydroxide = 0.310

Mole fraction = number of moles of a component ÷ total number of moles in the solution

Mole fraction = 0.310

In a saturaturated aqueous solution we have NaOH and water

∴ Number of moles of water molecules per unit = 1 - 0.310 = 0.690

However 0.690 moles of H₂O weighs = 0.690 mole × 18.01528 g/mol =12.43 g = 0.01243 kg of H₂O

But the molality = number of moles per Kilogram of H₂O

therefore molality of NaOH in the sample of solution =

(0.310 mol of NaOH )÷(0.01243 kg of H₂O)

= 24.93 mol/kg or ≅ 25.0 m

5 0

3 years ago

What is the molar solubility of AgCl in a 0.050 M NaCl solution? The Ksp of AgCl is 1.6 x 10-10. (Assume that the contribution o

Hitman42 [59]

The molar solubility of AgCl:

The molar solubility of AgCl in a 0.050 M NaCl solution is 8 x $10^{-8}$ M $Ag^{+}$

What is solubility?

The solubility is the quantity of reagent required to saturate the solution or bring about the dissociation reaction's equilibrium.

Reaction:

The dissociation reaction of AgCl in water is:

$AgCl$ ⇄ $Ag^{+} + Cl^{-}$

Each mole of AgCl that dissolves in this reaction yields 1 mole of both $Ag^{+}$ and $Cl^{-}$ . The concentration of either the Ag or Cl ions would then be equal to the solubility.

Solubility= [ $Ag^{+}$ ] = [ $Cl^{-}$ ]

Calculation:

in 0.050 M NaCl, the [ $Cl^{-}$ ] = 1 x $10^{-2}$

ksp = [ $Ag^{+}$ ] x [ $Cl^{-}$ ]

1.6 x $10^{-10}$ = [ $Ag{+}$ ] x ( 5 x $10^{-2}$ )

[ $Ag^{+}$ ] = 5 x $10^{+2}$ x 1.6 x $10^{-10}$

[ $Ag^{+}$ ] = 5 x 1.6 x $10^{-10+2}$

[ $Ag^{+}$ ] = 8 x $10^{-8}$ M

Learn more about molar solubility here,

brainly.com/question/13202097

#SPJ4

8 0

2 years ago

How many moles of potassium chloride (KCI) are in 252 mL of 0.33 mil/L KCI

Rom4ik [11]

Answer:

0.083 moles KCl

Explanation:

We have 252 mL of 0.33 M KCl. "M" is molarity, and it's measured in mol/L. Since we know the molarity is 0.33 mol/L, in order to find how many moles 252 mL is, we need to multiply our given volume by the molarity.

However, our given volume is in mL, so let's convert to L first by dividing 252 by 1000 (because there are 1000 mL in 1 L):

252/1000 = 0.252 L

Now, we can multiply 0.252 L by 0.33 mol/L to get moles since the L units will cancel out:

0.252 L * 0.33 mol/L = 0.08316

We have two significant figures here, so round 0.08316 to 0.083.

The answer is 0.083 moles.

Hope this helps!

7 0

3 years ago