Mg(OH)2 <br> SO2<br> (NH4)2SO4 <br> H3PO4 <br> (All the numbers are subscripts!)

Which natural polymers are involved with making clothing?

fredd [130]

Answer:

A. wool, silkworm, cocoon, and cellulose

Explanation:

I hope this helped!

7 0

3 years ago

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NH3 is an example of A. an element. B. an atom. C. a mixture. D. a compound.

Keith_Richards [23]

Answer:

I believe it's an example of a compound.

4 0

4 years ago

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An atom of gold has a mass of 3.271 x 10-22g. How many atoms of gold are in 5.00g of gold?

rosijanka [135]

To get the number of gold atoms, you have to divide the mass of the gold by the mass of the gold atom. It follows this simple equation $\frac{mass of gold}{mass of gold atom}$ .

Let x be the number of gold atoms. Plug in the values to a calculator.

x = $\frac{5.00g}{3.271 x 10-22g}$
Both have the same units so the unit gram(g) can be cancelled.
x then would be equal to 1.53x10^22. So there are 1.53x10^22 atoms of gold in 5 g of gold

3 0

4 years ago

when 6g acetic acid is dissolved in 1000cm3 of solution then how many molecules ionize out of 1000 acetic acid molecules

iVinArrow [24]

Answer:

24.8 molecules are ionized from 1000 acetic acid molecules.

Explanation:

Acetic acid, CH₃COOH dissociates in water, thus:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Ka = 6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]

<em>That means amount of CH₃COO⁻ (the dissociated form) that are produced is followed by the equilibrium of the weak acid.</em>

<em />

The initial molar concentration of acetic acid (Molar mass: 60g/mol) is:

6g ₓ (1mol / 60g) = 0.1 moles acetic acid, in 1000cm³ = 1L.

0.1 moles / L = <em>0.1M</em>

The 0.1M of acetic acid will dissociate producing X of CH₃COO⁻ and H⁺, thus:

[CH₃COOH] = 0.1M - X

[CH₃COO⁻] = X

[H⁺] = X

Replacing in Ka formula:

6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]

6.3x10⁻⁵ = [X] [X] / [0.1 - X]

6.3x10⁻⁶ - 6.3x10⁻⁵X = X²

6.3x10⁻⁶ - 6.3x10⁻⁵X - X² = 0

Solving for X

X = - 0.0025 → False solution, there is no negative concentrations.

X = 0.00248M

That means, a 0.1M of acetic acid produce:

[CH₃COO⁻] = X = 0.00248M solution of the ionized form.

In a basis of 1000 molecules:

1000 molecules × (0.00248M / 0.1M) = 24.8

<h3>24.8 molecules are ionized from 1000 acetic acid molecules.</h3>

8 0

3 years ago

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3 ( is ) . The equation for the

gayaneshka [121]

Answer: 25.89g of O2 (g)

Explanation:

We begin by finding the molar mass of compounds of interest in the given equation. 2 KClO3 ⟶ 2 KCl + 3 O 2

1 mole of KCLO3 = 39 .10 + 35.5 + (3x16) = 122.55g, 2Moles would be 2 x 122.5 = 245.1g

1 mole Oxygen gas = 32g, 3 moles would be, 3 x 32= 96g

According to the equation;

245.1g of KCLO3 produces 96g of O2 (g)

If 1g of KCLO3 produces 96÷245.1 of O2 (g) from heating.

66.1g of KCLO3 would produce; 96÷245.1 x 66.1 = 25.89g of O2 (g) from heating.

5 0

4 years ago

Mg(OH)2 SO2 (NH4)2SO4 H3PO4 (All the numbers are subscripts!)