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3 years ago

What would a substance that could undergo condensation polymerization most likely have?

Chemistry

2 answers:

AlekseyPX3 years ago

7 0

A carboxylic acid monomer and an amine monomer

earnstyle [38]3 years ago

3 0

Answer:

D Two alcohol functioning groups

Explanation:

edge test

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Which statement is true about the atoms in helium gas?

jasenka [17]

Answer: The answer is they are not closely packed

Explanation: HOPE THIS HELPS!!

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3 years ago

The leaves on a green plant soak up sunlight to make food. A cow's stomach digests grass that was eaten. These are examples of w

Nostrana [21]

Metabolism since it creates energy.

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2 years ago

Given the reactions below, (1) Na(s) + H2O(l) → NaOH(s) + 1/2 H2(g) LaTeX: \Delta H^o_{rxn}Δ H r x n o = −146 kJ (2) Na2SO4(s) +

slamgirl [31]

Answer:

ΔrxnH = -580.5 kJ

Explanation:

To solve this question we are going to help ourselves with Hess´s law.

Basically the strategy here is to work in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.

Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-

The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.

This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.

Taking 1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.

Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l) ΔrxnHº = 259 / 2 kJ 1/2 eq3

+ 2NaOH(s) + SO₃(g) ⇒ Na₂SO₄ (s) + H₂O (l) ΔrxnHº = -418 kJ - eq 2

+ 2Na (s) + 2 H₂O (l) ⇒ 2 NaOH (s) + H₂ (g) ΔrxnHº = -146 x 2 2 eq 1

Na₂O (s) + SO₃ (g) ⇒ Na₂SO₄ (s) ΔrxnHº = 259/2 + (-418) + (-146) x 2 kJ

ΔrxnH = -580.5 kJ

4 0

3 years ago

What are properties ?

frozen [14]

Answer:

Explanation:

In logic and philosophy, a property is a characteristic of an object; a red object is said to have the property of redness. The property may be considered a form of object in its own right, able to possess other properties.

8 0

2 years ago

If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of T

Simora [160]

Answer:

The new pH after adding HCl is 7.07

Explanation:

The formula for calculating pH of a buffer is

pH = pKa + log([Conjugate base]/[Acid])

Before adding HCl,

7.55 = 7.55 + log([Conjugate base]/[Acid])

⇔ log([Conjugate base]/[Acid]) = 0

⇔ [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M

⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol

After adding HCl (3.00 mL, 1.00 M)

⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)

New volume solution is 120 m L+ 3 mL = 123 mL

HCl is a strong acid, it will convert the conjugate base to acid form, or we can express

Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol

⇒ Concentration = 0.003/0.123 M

Mole of new Acid form = 0.006 + 0.003 = 0.009 mol

⇒ Concentration = 0.009/0.123 M

Use the formula

pH = pKa + log([Conjugate base]/[Acid])

= 7.55 + log(0.003 / 0.009) = 7.07

8 0

3 years ago