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Gre4nikov [31]
2 years ago
8

Purple Group Header Warm-Up Why can I smell a chocolate bar even though it is a solid? Read the description of the office myster

y and write your prediction below. On a warm afternoon, a Universal Space Agency office worker returned from lunch break and noticed the office smelled like chocolate. She didn't see anything in the air, but she noticed her coworker was opening the wrapper of a solid bar of chocolate. Why can the office worker smell the chocolate bar even though it is in solid form?
Chemistry
1 answer:
likoan [24]2 years ago
4 0

Answer:

The smell of a chocolate is from the presence of volatile compounds present in the chocolate bar which at room temperature readily changes phase from solid to liquid to vapor or gas

Explanation:

There are nearly 600 identified compounds present in a chocolate bar and out of these, there are volatile components which gives the chocolate bar its distinctive aroma.

These volatile chocolate contents readily change phase from solid to vapor, with very short duration liquid phase.

For example, 3 methylbutanal, vanillin, and several organic compounds which are known to be readily volatile.

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At 25 °C, an aqueous solution has an equilibrium concentration of 0.00253 M for a generic cation, A2+(aq), and 0.00506 M for a g
Sunny_sXe [5.5K]

Answer:

The equilibrium constant Ksp of the generic salt AB2 =  6.4777 *10^-8 M

Explanation:

Step 1: The balanced equation

AB2 ⇒ A2+ + 2B-

Step 2: Given data

Concentration of A2+ = 0.00253 M

Concentration of B- = 0.00506 M

Step 3: Calculate the equilibrium constant

Equilibrium constant Ksp of [AB2] = [A2+][B-]²

Ksp = 0.00253 * 0.00506² = 6.4777 *10^-8 M

The equilibrium constant Ksp of the generic salt AB2 =  6.4777 *10^-8 M

7 0
2 years ago
Read 2 more answers
What is the ratio of the diffusion rates of cl2 and o2? rate cl2 : o2 = 0.47 0.67 0.45 0.69 1.5?
matrenka [14]
The  ratio  of  the diffusion  rate  of Cl2  and   O2   is 1.5

    calculation
rate  of diffusion Cl2/ rate   of  diffusion  O2 =√  molar mass of Cl2/ molar mass  of O2

molar  mass of Cl2  = 35.5 x2 = 71  g/mol
molar  mass O2  = 16 x2 =32g/mol

that  is    rate  of diffusion Cl2/ rate  of  diffusion of O2  =√ 71/ 32=  1.5
7 0
2 years ago
Write a balanced half-reaction describing the oxidation of gaseous dihydrogen to aqueous hydrogen cations.
xz_007 [3.2K]
The oxidation state of hydrogen gas is 0 and oxidation state of hydrogen cation is +1.
There’s an increase in oxidation number therefore it’s an oxidation reaction.
Oxidation reactions give out electrons. The masses and charges on both sides should be balanced
Half reaction is
H2 —> 2H+ +2e
8 0
2 years ago
Read 2 more answers
A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is
Alik [6]
<span>Answer: A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en). contains 0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+ Kf for Cu(en)2^2+ is 1x10^20. so 1 Cu+2 & 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2 [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Cu+2 = 9.26 e-19 Molar since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
4 0
2 years ago
Gasoline has a density of 0.749 g/mL. There are 454 grams in a pound and 3.7854 litres in a gallon. How many pounds does 19.2 ga
alexgriva [62]

Answer:

119.9 pound

Explanation:

Given data:

Density of gasoline = 0.749 g/mL

Volume of gasoline = 19.2 gal (19.2× 37854 =72679.9 mL)

Mass = ?

Solution:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Now we will put the values in formula:

d = m/v

0.749 g/mL = m/ 72679.9 mL

m = 54437.25 g

gram to gallon:

54437.25/ 454

m = 119.9 pound

7 0
2 years ago
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