All categories

3 years ago

An experiment applies heat to a substance and produces a gas that is a new and different substance. Choose the true statement.

Chemistry

1 answer:

bearhunter [10]3 years ago

4 0

Well if the change is reversible then it is a physical change, but if not then it is a chemical change.

You might be interested in

How might the molecules inside an ice pack be different from the liquid inside of an insulated cup?

MatroZZZ [7]

Answer:

The molecules inside the ice pack are frozen therefore they are stuck in place. The molecules inside an insulated cup aren't frozen but still stuck inside of their container so that they can not fly out.

Explanation:

5 0

3 years ago

Give two uses of compound under each of the following, in medicine and in industry

NemiM [27]

Explanation:

#Medicine

Compounds used as medicines are most often organic compounds, which are often divided into the broad classes of small organic molecules (e.g., atorvastatin, fluticasone, clopidogrel) and "biologics" (infliximab, erythropoietin, insulin glargine),

#Industry

Polymers and plastics such as polyethylene, polypropylene, polyvinyl chloride, polyethylene terephthalate, polystyrene and polycarbonate

7 0

3 years ago

If a sample is thought to be several million years old, which method would best help to determine its absolute age?

Makovka662 [10]

The answer is B)Ar-40/k-40

6 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

2 years ago

How many sodium ions are in 1.4 kg of sodium chloride, NaCl?

Whitepunk [10]

Answer:

1.44 x 10²⁵ ions of Na⁺

Explanation:

Given parameters:

Mass of NaCl = 1.4kg = 1400g

Unknown:

Number of ions of sodium = ?

Solution:

The compound NaCl in ionic form can be written as;

NaCl → Na⁺ + Cl⁻

In 1 mole of NaCl we have 1 mole of sodium ions

Now, let us find the number of moles in NaCl;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Number of moles = $\frac{1400}{58.5}$ = 23.93mol

So;

Since 1 mole of NaCl gives 1 mole of Na⁺

In 23.93 mole of NaCl will give 23.93 mole of Na⁺

1 mole of a substance = 6.02 x 10²³ ions of a substance

23.93 mole of a substance = 6.02 x 10²³ x 23.93

= 1.44 x 10²⁵ ions of Na⁺

7 0

2 years ago