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3 years ago

Open Ended

Chemistry

1 answer:

kakasveta [241]3 years ago

4 0

Answer:

Explanation:

21. Atoms are not created or destroyed means that atoms that you begin with are the atoms that you will end with. The catch is that the atoms will rearrange to give you new compounds, but the atoms that you initially had are the atoms you will still have after reaction. For eg, if you started with eggs and made omelet. Omelet is a "new" compound, but the atoms that were in the eggs have rearranged to become the omelet so can you see that atoms were not created or destroyed to make the omelet.

22. Yes because amount of products you make depends on how much reactants you have. For eg, I need two graham cracker(GC), one marshmallow(M), and one chocolate (C) to make a s'more. If I get more of each item then I can make more s'mores and consequently having minimum amounts results in less s'mores that I make.

23. Not possible, due to law of conservation of matter and energy. Atoms cannot be created nor be destroyed, they are simply rearranged. For eg, Taking A + B cannot give you a new compound with a chemical formula D or XZ. A + B can however give you AB which is rearrangement of the starting atoms.

24. Chemical equation is balanced when atoms on reactant side and atoms of product side are in equal counts. I have attached a graphic below for more help.

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The number of grams of H2 in 1470 mL of H2 gas.

Lorico [155]

Answer:

0.1313 g.

Explanation:

It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.

Suppose that hydrogen behaves ideally and at STP conditions.

Using cross multiplication:

1.0 mol of hydrogen occupies → 22.4 L.

??? mol of hydrogen occupies → 1.47 L.

∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.

Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:

The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen = (6.563 x 10⁻² mol)(2.0 g/mol) = 0.1313 g.

5 0

3 years ago

A tetraphenyl phosphonium chloride (TPPCl) powder (FW=342.39) is 94.0 percent pure. How many grams are needed to prepare 0.45 L

slega [8]

Answer:

5.41 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For tetraphenyl phosphonium chloride :

Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)

Volume = 0.45 L

Thus, moles of tetraphenyl phosphonium chloride :

$Moles=0.033 \times {0.45}\ moles$

Moles of TPPCl = 0.01485 moles

Molar mass of TPPCl = 342.39 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.01485\ g= \frac{Mass}{342.39\ g/mol}$

Mass of TPPCl = 5.0845 g

Also,

TPPCl is 94.0 % pure.

It means that 94.0 g is present in 100 g of powder

5.0845 g is present in 5.41 g of the powder.

Answer - 5.41 g

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3 years ago

Define a transverse wave

Mumz [18]

Transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.

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3 years ago

Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc

svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ, ΔSºrxn = 0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG = 1.8 x 10⁴ J

K = 8.2 x 10⁻²

Explanation:

a) C6H5−CH2CH3 ⇒ C6H5−CH=CH2 + H₂

ΔHf kJ/mol -12.5 103.8 0

ΔGºf kJ/K 119.7 202.5 0

Sº J/mol 255 238 130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate ΔHºrxn we have

ΔHºrxn = ΔHfº C6H5−CH=CH2 + ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn = 103.8 kJ + 0 kJ - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 + ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn= 202.5 kJ + 0 kJ - 119.7 kJ = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn = ΔHrxn -TΔS

we see that will happen when the term TΔS becomes greater than ΔHrxn since ΔS is positive , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that ΔºHrxn and ΔSºrxn remain constant at the higher temperature and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K = (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

ΔGrxn = ΔHrxn -TΔS

to calculate ΔGrxn with the assumption that ΔHº and ΔSºremain constant.

ΔG = 116.3 kJ - (600+273 K) x 0.113 kJ/K = 116.3 kJ - 873 K x 0.113 kJ/K

ΔG = 116.3 kJ - 98.6 kJ = 17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K x 873 K)) = 8.2 x 10⁻²

8 0

3 years ago