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3 years ago

Both acids and bases produce ions when dissolved in water. Propose a way to describe what

Physics

2 answers:

Ber [7]3 years ago

7 0

What he said I just need points and I wanna look hella smart

EleoNora [17]3 years ago

6 0

Answer:

See explanation

Explanation:

Acids and bases contain ions that interact with water. According to the Arrhenius definition, acids are substances that produce hydrogen ion in water while bases are substances that produce hydroxide ion in water.

The pH scale is a graphic description of the hydrogen or hydroxide ion present in a sample. Since pH= -log[H^+], the higher the pH , the lower the hydrogen ion concentration and vice versa.

Similarly, pOH= -log [OH^-] , hence the more the OH^- concentration the lower the pOH.

However pH + pOH =14.

Thus the concentration of hydrogen or hydroxide ions present determines the pH of any solution.

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It's downward vector from the centre of apple in the direction to the centre of the Earth.

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3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

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3 years ago

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9 atoms per molecule.

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3 years ago

Read 2 more answers

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frutty [35]

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3 years ago

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MissTica

Answer: