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mezya [45]
3 years ago
7

What makes slime, slime?

Chemistry
2 answers:
Serggg [28]3 years ago
8 0

Answer:

Glue,water,borax,food coloring,

Explanation:

Naya [18.7K]3 years ago
6 0
Slime flows like a liquid, but unlike familiar liquids (e.g., oil, water), its ability to flow, or viscosity, is not constant. So it's a fluid, but not a regular liquid. Scientists call a material that changes viscosity a non-Newtonian fluid. The technical explanation is that slime is a fluid that changes its ability to resist deformation according to shear or tensile stress.

What this means is, when you pour slime or let it ooze through your fingers, it has a low viscosity and flows like a thick liquid. When you squeeze a non-Newtonian slime, like oobleck, or pound it with your fist, it feels hard, like a wet solid. This is because applying stress squeezes the particles in the slime together, making it hard for them to slide against each other.

Most types of slime are also examples of polymers. Polymers are molecules made by linking together chains of subunits.


The specifics of how a type of slime works depends on its chemical composition, but the basic explanation is that chemicals are mixed to form polymers. The polymers act as a net, with molecules sliding against each other.


Two solutions are combined to make classic slime. One is diluted school glue, or polyvinyl alcohol in water. The other solution is borax (Na2B4O7.10H2O) in water.
Borax dissolves in water into sodium ions, Na+, and tetraborate ions.
The tetraborate ions react with water to produce the OH- ion and boric acid:
B4O72-(aq) + 7 H2O <—> 4 H3BO3(aq) + 2 OH-(aq)
Boric acid reacts with water to form borate ions:
H3BO3(aq) + 2 H2O <— > B(OH)4-(aq) + H3O+(aq)
Hydrogen bonds form between the borate ion and the OH groups of the polyvinyl alcohol molecules from the glue, linking them together to form a new polymer: slime.
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How many grams of H2SO4 are needed to prepare 500. mL of a .250M solution?
zavuch27 [327]

Answer:

We need 12.26 grams H2SO4

Explanation:

Step 1: Data given

Volume of a H2SO4 solution = 500 mL = 0.500 L

Concentration of the H2SO4 solution = 0.250 M

Molar mass of H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = concentration * volume

Moles H2SO4 = 0.250 M * 0.500 L

Moles H2SO4 = 0.125 moles

Step 3: Calculate mass of H2SO4

Mass of H2SO4 = moles * molar mass

Mass of H2SO4 = 0.125 moles * 98.08 g/mol

Mass of H2SO4 = 12.26 grams

We need 12.26 grams H2SO4

7 0
3 years ago
How might the yield of 1-bromobutane be affected if water was not added, and what product(s) would be favored?
jek_recluse [69]
  • Due to the inability of the reaction to take place, the yield of 1-Bromobutane would drop.
  • Since 1-Butanol won't react with the additional sodium bromide, bromination won't happen.
  • If water had been supplied, the equilibrium would have shifted extremely far to the left, preventing the reactants from interacting with the acid and favoring the yield of 1-bromobutane instead.
<h3>What is Bromination?</h3>
  • When a substance undergoes bromination, bromine is added to the compound as a result of the chemical reaction.
  • After bromination, the result will have different properties from the initial reactant.
  • For example, an alkene is brominated by electrophilic addition of Br_{2}.
  • Benzene ring bromination by electrophilic aromatic substitution.

Learn more about Bromine here:

brainly.com/question/862562

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1 year ago
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Nina [5.8K]
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If you dilute 19.0 mL of the stock solution to a final volume of 0.310 L , what will be the concentration of the diluted solutio
Dahasolnce [82]

Answer:

M_2=0.613M_1

Explanation:

M_1 = Concentration of stock solution

M_2 = Concentration of solution

V_1 = Volume of stock solution = 19 mL

V_2 = Volume of solution = 0.31 L= 310 mL

We have the relation

M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{M_119}{310}\\\Rightarrow M_2=M_1\times\dfrac{19}{310}\\\Rightarrow M_2=0.613M_1

\boldsymbol{\therefore M_2=0.613M_1}

The concentration of the diluted solution will be 0.613 times the concentration of the stock solution.

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