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LenKa [72]
3 years ago
14

Using knowledge of intermolecular forces, explain why the yellow food coloring mixed with the water and the acetone was able to

remove the blue dye from the glitter.
(Basically the experiment was we mixed yellow food coloring with acetone and water and added it to glitter, and it removed the blue dye from the glitter. Please mention intermolecular forces in your answer.)
Chemistry
1 answer:
Papessa [141]3 years ago
8 0

The yellow coloring is a polar substance hence it dissolve in water. Blue dye is nonpolar hence it dissolves in less polar acetone.

In chemistry, there is a general rule that like dissolves like. A substance can only dissolve in another if there is some kind of intermolecular interaction between the substances.

The yellow food coloring is a polar substance hence it is able to dissolve in water which is a polar solvent. The blue dye is a nonpolar substance hence it dissolve in acetone which is a less polar solvent.

Learn more:  brainly.com/question/9847214

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Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 7.80 moles of magnesium perchlorate, Mg(ClO4)2.
Dmitry [639]
 The number of moles of moles of  Magnesium,chlorine and oxygen atoms in 7.80 moles of Mg(ClO4)2 is calculated as below

find the total number of each atom in Mg(ClO4)2

that is mg = 1 atom
        Cl =  1x2 = 2 atoms
        O  = 4  x2 = 8 atoms

then multiply 7.80 moles with total number of each atom , to get the number moles of each atom
that is

Mg = 7.80 x1= 7.80  moles
cl =   7.80  x2=15.6  moles
O =  7.80 x8= 62.4 moles

5 0
3 years ago
URGENT CHEMISTRY EXPERT!
vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl

(b) Moles of CuCl₂

\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} =  \text{3.340 mmol CuCl}_{2}

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

\text{Moles of Na$_{3}$PO}_{4} =  \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}

(d) Volume of Na₃PO₄

V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}

Part 2. Net ionic equation

(a) Molecular equation

\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>  

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0
3 years ago
How do you convert between the mass and the number of moles of a substance?
Ksivusya [100]
To convert from grams to moles, you divide the number of grams by the molar mass To convert  from moles to grams, you multiply by the molar mass. You must first calculate the molar mass of the substance
6 0
3 years ago
If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?
Hitman42 [59]
Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.
4 0
3 years ago
I need chemistry help text me at (951) 897-8325 PLEASE
user100 [1]
Just post the question on here
7 0
2 years ago
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