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3 years ago

Why is protein classified as a polymer?

Chemistry

2 answers:

Anika [276]3 years ago

3 0

Its made up of a long chain of monomers so proteins are polymers made of amino acids I hope this helps! ^^;

Reil [10]3 years ago

3 0

Definitley a protine is counted as a polymer

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Do you know how to do this? if you have 4 grams of carbon-14, how much carbon 14 remains after 11,460 years? the half life is 5,

Delvig [45]

Carbon dating has given archeologists a more accurate method by which they can determine the age of ancient artifacts. The halflife of carbon 14 is 5730 ± 30 years, and the method of dating lies in trying to determine how much carbon 14 (the radioactive isotope of carbon) is present in the artifact and comparing it to levels

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3 years ago

Given 7.40 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100

Dafna11 [192]

On complete conversion (100% yield) 9.75 g of ethyl butyrate will be produced. Below is the solution.....

4 0

3 years ago

(First to answer gets some good) Order the following components of the universe from largest to smallest: Moon, planet,universe,

cluponka [151]

Answer:

Universe, galaxy, solar system, star, planet, moon and asteroid.

Explanation:

You're welcome!

7 0

3 years ago

What is the volume of 40.0 grams of argon gas at STP ?

MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Aqueous Solutions

States of Matter

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 40.0 g Ar

[Solve] L Ar

Step 2: Identify Conversions

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

Step 3: Convert

[DA] Set up: $\displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 24.9235 \ L \ Ar$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

24.9235 L Ar ≈ 24.9 L Ar

5 0

2 years ago

Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super

dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below: $X^{m+}Y^{n-}\rightarrow X_nY_m$

Thus, for all the given combinations, we obtain:

- Y⁻

$X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3$

- Y²⁻

$X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3$

- Y³⁻

$X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY$

Best regards!

8 0

2 years ago