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Dima020 [189]
3 years ago
6

How does energy change matter?

Chemistry
2 answers:
wlad13 [49]3 years ago
7 0
Adding or removing energy from matter causes a physical change as matter moves from one state to another.

Example : adding thermal energy to liquid water causes it to become steam or vapor ( a gas ) and removing energy from liquid water causes it to become ice (a solid).
Aleks [24]3 years ago
3 0

Answer:

Adding or removing energy from matter causes a physical change as matter moves from one state to another. For example, adding thermal energy (heat) to liquid water causes it to become steam or vapor (a gas). And removing energy from liquid water causes it to become ice (a solid).

Explanation:

Your welcome

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State the Third Law of Motion on the basis of the Third law explain the following [a] A boat tend to leave the shore when passen
Firlakuza [10]
C because it has to do with opposite motions and stuff
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Drag the tiles to the correct boxes. Not all tiles will be used.
gizmo_the_mogwai [7]

Answer:

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Explanation:

I have matched correctly. Just check SI units and there measures.

4 0
2 years ago
What steps should you take to respond to an accident?
Marina86 [1]
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5 0
2 years ago
What is the metal cation in K,SO,? (cation is pronounced cat-ion and refers to a positively charged ion)
slamgirl [31]

Answer:

Potassium cation = K⁺²

Explanation:

The metal cation in K₂SO₄ is K⁺². While the anion is SO₄²⁻.

All the metals have tendency to lose the electrons and form cation. In given compound the metal is potassium so it should form the cation. The overall compound is neutral.

The charge on sulfate is -2. While the oxidation state of potassium is +1. So in order to make compound overall neutral there should be two potassium cation so that potassium becomes +2 and cancel the -2 charge on sulfate and make the charge on compound zero.

2K⁺²  ,  SO₄²⁻

K₂SO₄

8 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
3 years ago
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