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Rom4ik [11]
3 years ago
13

A student decides to dissolve crackers in their soup as fast as possible. They must first increase the surface area of the crack

ers. How could this be done? a.Break the crackers into pieces b.Pile the crackers on top of each other c.Wrap the crackers in plastic to keep them fresh d.Put crackers in the bottom of the bowl
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
8 0

Answer: The correct option is A ( Break the crackers into pieces.)

Explanation:

The rate of reaction between two substances can be affected by the following factors:

--> surface area of reacting substances

--> concentration

--> reaction mechanism

--> temperature

--> the presence of a catalyst and

--> pressure ( if gaseous)

Reagents in solution tend to react faster than reagents in solid form. This is because molecules are freer to move about and collide more effectively in solutions. For example, a lump of zinc will not react as fast as powdered zinc with dilute hydrochloric acid to produce hydrogen. This is the same with the reaction between the crackers and soup. To increase the reaction of the substances, the crackers needs to be broken into pieces(thus,increased surface area)

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which equation is setup correctly to determine the volume of a 1.5 mol sample of oxygen gas at 22 degrees Celsius and 100 kPa
rjkz [21]
Hello!

We have the following data:


v (volume) = ? (in L)
n (number of mols) = 1,5 mol
T (temperature) = 22 ºC 
First let's convert the temperature on the Kelvin scale, let's see:
TK = TºC + 273,15
TK = 22 + 273,15
TK = 295,15

P (pressure) = 100 kPa → P = 100000 Pa → P ≈ 0,987 atm
R (gas constant) = 0,082 atm.L / mol.K

<span>We apply the data above to the Clapeyron equation (gas equation), let's see:

</span>P*V = n*R*T

0,987*V = 1,5*0,082*295,15

0,987V = 36,30345

V =  \dfrac{36,30345}{0,987}

\boxed{\boxed{V \approx 36,78\:L}} \end{array}}\qquad\checkmark

I hope this helps. =)
4 0
4 years ago
Sulfonation of naphthalene, C10H8, results in two products. One product is kinetically favored and predominates in the beginning
Marrrta [24]

Explanation:

The sulfonation of the naphthalene yield 2 products under different conditions:

<u>When the reaction is carried at 80 °C, 1-naphthalenesulfonic acid is the major product because it is kinetically favoured product as arenium ion formed in the transition state corresponding to 1-naphthalenesulfonic acid is more stable due to better resonance stabilization. </u>

<u>When the reaction is carried at 160 °C, 2-naphthalenesulfonic acid is the major product as it is more stable than 1-naphthalenesulfonic acid because of  steric interaction of the sulfonic acid group in 1-position and the hydrogen in 8-position.</u>

The products are shown in image below.

7 0
4 years ago
A standard solution is prepared for the analysis of fluoxymesterone (C20H29FO3), an anabolic steroid. A stock solution is first
VLD [36.1K]

Answer:

The final solution has a concentration of 5.95 * 10^-6 M

Explanation:

<u>Step 1:</u> data given

Mass of fluoxymesterone : 10mg = 0.01g

Molar mass of  fluoxymesterone = 336g/mole

Volume = 500 mL =0.5 L

<u>Step 2:</u> calculate Concentration of  fluoxymesterone

Concentration = mole / volume

⇒ To find the concentration, we first need to find the number of moles

⇒moles fluoxymesterone= mass of fluoxymesterone / Molar mass

0.01g / (336g/mole) = 2.976 *10^-5 mole

C =  2.976 *10^-5 mole / 0.5 L = <u>5.95 * 10 ^-5 M</u>

<u>Step 3:</u> Calculate dilution

⇒ If 10 mL  is diluted to 100 ml then this is a 10x dilution (100 ml /10 ml = 10) so the final solution will be 10 x more dilute.

⇒5.95 * 10^-5 M /10  = 5.95 ^* 10^-6 M

We can control this through the following formula: M1 * V1 = M2 * V2

5.95 *10^-5 * 10 * 10 *^-3 = M2 *0.1 L

M2 = 5.95 * 10^-6 M

6 0
4 years ago
The ---------- is directly above the troposphere
Debora [2.8K]
This would be the stratosphere. which is the second layer of the atmosphere.
7 0
3 years ago
Help me with number 11,14,15,16 please
elena55 [62]
11. The principle of conservation mass is that in any close system subjected to no external forces and mass is constant of its change of form.
5 0
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