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nydimaria [60]
2 years ago
9

Position (m)

Chemistry
1 answer:
Alex777 [14]2 years ago
8 0

Option C. The object is returning to the start at a constant speed.

<h3>Data points of the Position vs Time graph</h3>

The following data points will be used to determine the motion of the object.

<u>Position               Time</u>

12                          4

10                          6

2                            8

0                           10

From the data above, the position of the object is decreasing towards zero or start point.

Thus, the object is returning to the start at a constant speed.

Learn more about position here: brainly.com/question/2364404

#SPJ1

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Calculate the amount of heat water absorbs from a piece of hot metal using the following data: 75.0 g of cold water is placed in
AysviL [449]

Answer:

Amount of heat absorbed by water is 2604.54 J.

Explanation:

Amount of heat absorbed by water = m_{water}\times C_{water}\times \Delta T_{water}

where m represents mass, C represents specific heat and \Delta T represents change in temperature.

Here m_{water}=75.0 g , C_{water}=4.184J/(g.^{0}\textrm{C}) and \Delta T = (final temperature - initial temperature) = (29.5-21.2) ^{0}\textrm{C} = 8.3 ^{0}\textrm{C}

So, amount of heat heat absorbed by water

     = (75.0g)\times (4.184\frac{J}{g.^{0}\textrm{C}})\times (8.3^{0}\textrm{C})

     = 2604.54 J

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Which of the following is an electrolyte?<br><br> a. PbCl2<br> b. MgS<br> c. HNO3<br> d. Al2(Cr2O7)3
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3 years ago
What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250
Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons \text{H}^{+} than hydroxide ions \text{OH}^{-}. The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that x \; \text{L} of the 0.307 \text{mol} \cdot \text{dm}^{-3} sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, x is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain 0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x moles of acetic acid and 0.307 \; x moles of acetate ions.

Let \text{HAc} denotes an acetic acid molecule and \text{Ac}^{-} denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of 10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}. There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}

By definition:

\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)

The question states that

\text{K}_{a} = 1.75 \times 10^{-5}

such that

10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241

Thus it takes 0.0241 \; \text{L} of sodium hydroxide to produce this buffer solution.

6 0
3 years ago
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