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erma4kov [3.2K]
3 years ago
14

Two teams of students are having a tug-of-war. Team 1 pulls with a force of 150 newtons (N) while team 2 pulls with a force of 2

00 newtons (N). Based on this information, which team would be predicted to win the tug-of-war?
Chemistry
1 answer:
Elena-2011 [213]3 years ago
7 0

Answer:

Team 2

Explanation:

Team 1 pulls with a frce of 150N

Team 2 pulls with a force of 200N

In a tug-of-war, the goal is to pull the other team over a certain distance and this is determined by the force applied.

If the two team pulls with equal force, there wold be no net force and the teams stays in the same position.

In this case however, team 2 pulls with a force greater than team 1. There is a net force of 50N in favor of Team 2. This means team 2's force would neutralize that of team 1 and the excess force would be use in pulling Team 1

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Which of the following describes an ionic bond?
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The distance between Earth and its moon is 384000000 meters. Express this distance in kilometers.
Andreyy89
384000000m

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3 years ago
Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th
Tpy6a [65]

pH = 13.5

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}

The mixture would contain

  • 0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol} of \text{OH}^{-} and
  • 0.1 \times 0.5 = 0.05 \; \text{mol} of \text{Ac}^{-}

if \text{Ac}^{-} undergoes no hydrolysis; the solution is of volume 0.1 + 0.4 = 0.5 \; \text{L} after the mixing. The two species would thus be of concentration 0.30 \; \text{mol} \cdot \text{L}^{-1} and 0.10 \; \text{mol} \cdot \text{L}^{-1}, respectively.

Construct a RICE table for the hydrolysis of \text{Ac}^{-} under a basic aqueous environment (with a negligible hydronium concentration.)

\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}

The question supplied the <em>acid</em> dissociation constant pK_afor acetic acid \text{HAc}; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant pK_b for its conjugate base, \text{Ac}^{-}. The following relationship relates the two quantities:

pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})

... where the water self-ionization constant pK_w \approx 14 under standard conditions. Thus pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3. By the definition of pK_b:

[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b =  10^{-pK_{b}}

x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}

x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}

[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}

pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5

6 0
3 years ago
Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the equation:
weqwewe [10]

Answer:

D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117M

Explanation:

Based on the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

And knowing:

Kc = [PCl₃] [Cl₂] / [PCl₅] = 1.80

When you add PCl₅ into a flask, this gas will react producing PCl₃ and Cl₂ until [PCl₃] [Cl₂] / [PCl₅] = 1.80

This could be written as:

[PCl₃] = X

[Cl₂] = X

[PCl₅] = 0.125M - X

<em>Where X represents the moles of PCl₅ that react, </em><em>reaction coordinate.</em>

Replacing in Kc expression:

[PCl₃] [Cl₂] / [PCl₅] = 1.80

[X [X] / [0.125 - X] = 1.80

X² = 0.225 - 1.80X

0 = -X² -1.80X + 0.225

Solving for X:

X = -1.9M → False solution, there is no negative concentrations

X = 0.11735M → Right solution.

Replacing, concentrations in equilibrium are:

[PCl₃] = X

[Cl₂] = X

[PCl₅] = 0.125M - X

[PCl₃] = 0.117M

[Cl₂] = 0.117M

[PCl₅] = 0.00765M

And right option is:

<h3>D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117M</h3>
7 0
3 years ago
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