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3 years ago

Do you think creativity can contribute to scientific inquiry and investigation?

1 answer:

3 0

Yes. It is only through creativity that scientific inquiry exists. Someone had the wacky creative idea that there should be something that is so strong, and inflammable, and just generally amazing like a super... thing! And tadaaah, we have Kevlar. The stuff in bullet proof vests.

You might be interested in

In a chemical reaction, the mass of the reactants always equals the mass of the products.

Viefleur [7K]

Answer: true

Explanation: according to the law of conservation of mass the mass of reactants would equal the mass of the products

4 0

3 years ago

Which of the following atoms has the highest ionization energy?<br> Na, Cs, K, Li

IceJOKER [234]

Li is the highest ionization energy

7 0

3 years ago

Are the cells in this image prokaryotic or eukaryotic

Bas_tet [7]

Answer:

Prokaryotic is the answer!

Explanation:

I know this because, the nucleolus is absent in the image.

PLZ MARK AS BRAINLIEST! HOPE THIS HELPS! :)

ps: i have the same profile photo lol, love it!

3 0

3 years ago

A 10.00 mL sample of vinegar (an aqueous solution of acetic acid) is titrated with 0.5062 M NaOH(aq) and 16.58 mL is required to

sammy [17]

Answer:

5.01%

Explanation:

Density of vinegar = mass/volume

Mass of 10.00 mL = density x volume

= 1.006 x 10 = 10.06 g

From the equation of reaction:

$CH_3COOH(aq)+NaOH(aq)-->CH_3COONa(aq)+H_2O(l)$

1 mole pf CH3COOH requires 1 mole of NaOH for neutralization.

mole of NaOH = molarity x volume

= 0.5062 x 0.01658

= 0.008392796‬ mole

0.008392796‬ mole of NaOH will therefore require 0.008392796‬ mole of CH3COOH.

mass of CH3COOH = mole x molar mass

= 0.008392796‬ x 60.052

= 0.504 g

Percentage by mass of acetic acid in the vinegar = 0.504/10.06 x 100%

= 5.01%

The percent by mass of acetic acid in the vinegar is 5.01%

3 0

3 years ago

Calculate the percent composition by mass of the following compounds that are important starting materials for synthetic polymer

notsponge [240]

Answer:

C₃H₄O₂ → 50% C; 5.5 % H; 44.5% O

C₄H₆O₂ → 56 % of C; 7 % of H; 37% of O

C₃H₃N → 68 % of C; 6 % of H; 26 % of N

Explanation:

We determine the molar mass of each compound:

C₃H₄O₂ → 72 g/mol

In 1 mol of acrylic acid (72 g), we have:

3 moles of C → 12 g/mol . 3 mol = 36 g of C

4 moles of H → 1 g/mol . 4 mol = 4 g of H

2 moles of O → 16g/mol . 2 mol = 32 g of O

Then in 100 g of salt, we may have:

100 . 36 / 72 = 50 % of C

4 . 100 / 72 = 5.5 % of H

32 . 100 / 72 = 44.5 % of O

C₄H₆O₂ → 86 g/mol

In 1 mol of methyl acrylate (86 g), we have:

4 moles of C → 12 g/mol . 4 mol = 48 g of C

6 moles of H → 1 g/mol . 6 mol = 6 g of H

2 moles of O → 16g/mol . 2 mol = 32 g of O

Then in 100 g of salt, we may have:

100 . 48 / 86 = 56 % of C

6 . 100 / 86 = 7 % of H

32 . 100 / 86 = 37 % of O

C₃H₃N → 53 g/mol

In 1 mol of acrylonitrile (53 g), we have:

3 moles of C → 12 g/mol . 3 mol = 36 g of C

3 moles of H → 1 g/mol . 3 mol = 3 g of H

1 mol of N → 14g/mol . 1 mol = 14 g of N

Then in 100 g of salt, we may have:

100 . 36 / 53 = 68 % of C

3 . 100 / 53 = 6 % of H

14 . 100 / 53 = 26 % of N

3 0

3 years ago