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GuDViN [60]
3 years ago
10

Define force and speed​

Chemistry
2 answers:
nalin [4]3 years ago
8 0

Answer:

Force is a push or pull which changes or tend to change the position of a body.

The rate of change its position with time or magnitude is called speed.

Stells [14]3 years ago
7 0

<em>Force</em><em> </em><em>=</em><em> </em><em>The</em><em> </em><em>external</em><em> </em><em>energy</em><em> </em><em>that</em><em> </em><em>changes</em><em> </em><em>or</em><em> </em><em>tends</em><em> </em><em>to</em><em> </em><em>change</em><em> </em><em>the</em><em> </em><em>state</em><em> </em><em>of</em><em> </em><em>any</em><em> </em><em>body</em><em> </em><em>or</em><em> </em><em>object</em><em> </em><em>is</em><em> </em><em>called</em><em> </em><em>force</em><em>.</em>

<em>Speed</em><em> </em><em>=</em><em> </em><em>The</em><em> </em><em>rate</em><em> </em><em>of</em><em> </em><em>distance</em><em> </em><em>is</em><em> </em><em>called</em><em> </em><em>speed</em><em>.</em>

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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

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