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Akimi4 [234]
3 years ago
12

Summarize your findings in a short report of 150 words. Include your hypothesis, observations, data, calculations, and conclusio

n. The answers to the questions below should be included in your conclusion.
Why didn't the box slide off the table before you put enough coins in the bag?
What was the coefficient for the smooth surface? For the rough surface?
What was the difference between the two coefficients? What caused this difference?
What factors caused a margin for human error in this investigation?
Would the coefficient vary if you tied the string to a similar object of different masses, but on the same surface? Explain.
What did you learn from this investigation? Be thoughtful in your answer.
Chemistry
1 answer:
Damm [24]3 years ago
5 0

Answer:

Follow these steps.

1. Fill the matchbox with pebbles. Weigh the matchbox with the pebbles inside. Record that weight.

2. Tie the string to the box. Allow the string to hang over the edge of the table.

3. Tie the other end of the string to a corner of the plastic bag, leaving an opening to put in coins.

4. Add coins one by one until the box is pulled off the table.

5. Count and record the number of coins and the weight of the bag with the coins in it.

6. Lay the round sticks on the table about 1 inch apart and about 2 inches from the edge of the table.

7. Put the matchbox on the rollers farthest from the edge of the table.

8. Now add coins one by one to the bag until the box is pulled off the table.

9. Count and record the number of coins and the weight of the bag with the coins in it.

10. Repeat the experiment. Determine your margin of error if your results vary. For accuracy, repeat the experiment if desired.

11. Using the equation for the coefficient of friction in the text above, determine the coefficient of friction for the matchbox in each experiment. Include this data in your summary.

Explanation:

I think this is useful

please make me as breainlest

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Exocytosis : exo- means outside , -cyte means cell

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3 years ago
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(vi) A solvent extraction in which two ...... liquids move in opposite directions in
ad-work [718]

Answer:

(A). Immmiscible

Liquid/Liquid extraction is a solvent extraction in which two immiscible liquids move in opposite directions are in continuous contact with each other.

Explanation:

A liquid/liquid extraction is a method used to separate mixtures based on relative solubilities in two immiscible liquids, i.e., liquids that do not dissolve in each other but rather form separate layers when placed in the same glassware.

The immiscibility of the liquids is as a result of the two liquids having different polarities. One liquid is polar while the other is non polar, usually water which is polar and an organic solvent which is non-polar. The most common pair of extraction solvents used is diethyl ether or simply ether and water. They separate into layers and the position most each solvent in either the bottom or top layer is determined by its density. Ether is always found on the top layer while water is found innthe bottom layer because ether is less than than water. The density of ether is 0.713 g/cm3 and the density of H2O is 1.0 g/cm3.

The compound to be extracted dissolves into either of the layers based on the principle that like dissolves like. Polar compounds will dissolve into the water layer while organic nor non-polar compounds will dissolve in the ether layer. The compound is the extracted from either of the layer it is dissolved in.

4 0
3 years ago
Estimate the solubility of CaCO3
Fudgin [204]

Answer:

CaCO3(s) <==> Ca2+(aq) + CO3 2-(aq)

Let X = the amount of CaCO3 dissolved in Na2CO3 or the amount of each ions formed i.e. Ca2+ and CO3 2–. Then, put this into the formula :

Ksp = [Ca2+] [CO3 2-]

5 × 10^-9 = [X]*[X]

X^2 = 5 × 10^-9

X = √5 × 10^-9

X = 7.07 x 10^-5 mole/L

So, the amount of CaCO3 dissolved in 0.1M Na2CO3 or the amount of Ca2+ and CO3 2- ions produced is 7.07 x 10^-5 mole/L

4 0
3 years ago
A 5.00g piece of metal is heated to 100.0°C, then placed in a beaker containing 20.0 of water at 10.0°C. The temperature of the
Ahat [919]
Answer:

This metal has a specific heat of 0.9845J/ g °C

Explanation:
Step 1: Given data
q = m*ΔT *Cp
⇒with m = mass of the substance
⇒with ΔT = change in temp = final temperature T2 - initial temperature T1
⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = TO BE DETERMINED)

Step 2: Calculate specific heat
For this situation : we get for q = m*ΔT *Cp
q(lost, metal) = q(gained, water)

- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)
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Cpmetal = 418.4 / (-5*-85) = 0.9845 J/g °C

This metal has a specific heat of 0.9845J/ g °C
7 0
3 years ago
For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the
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Answer is: activation energy of this reaction is 212,01975 kJ/mol.<span>
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k</span>₁ = 0,000643 1/s.<span>
k</span>₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

<span> 1/T</span>₁ = 1/622 K = 0,0016 1/K.<span>
1/T</span>₂ = 1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>

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