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Musya8 [376]
2 years ago
10

the energy to break 1 mol of C-C bonds is 348 kj/mol. what would be the minimum frequency of a single photon that would break a

single c-c bond
Chemistry
1 answer:
mr_godi [17]2 years ago
4 0

Answer:

f = 5.25 x 10³⁸ s⁻¹

Explanation:

Energy = 348KJ = 348,000 J

Frequency = ?

Energy and frequency are related by the equation below;

E = hf

where h = Planck's constant =  (6.626 x 10-34 J · s),

Upon making f subject of formular;

f = E / h

substituting the values, we have;

f = 348,000 / 6.626 x 10-34

f = 52,520 x 10 ³⁴ s⁻¹

f = 5.25 x 10³⁸ s⁻¹

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Using the mass of the proton 1.0073 amu and assuming its diameter is 1.0×10−15m, calculate the density of a proton in g/cm3.
icang [17]

Answer : 3.2 X 10^{15} g/cm^{3}

Explanation :  To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = 1.66054 X 10^{-24} g

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X  1.66054 X 10^{-24} g/amu = 1.673 X 10^{-24} g

Now, Volume = 1/6πd³ as diameter is given as 1.0 X 10^{-15} m converting it to cm will require to multiply with 100

∴ Volume  = 1/6π (1.0 X 10^{-15}mX 100 cm / 1 m)^{3}

Hence, volume =  5.236 X 10^{-40} cm^{3}

Therefore, Density = mass / volume

∴ Density =  1.673 X 10^{-24} g / 5.236 X 10^{-40} cm^{3}

Therefore, Density will be 3.2 X 10^{15} g/cm^{3}.

6 0
2 years ago
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A mass spectrometer has determined the mass and abundances of all isotopes of an unknown element. The first isotope has a mass o
shepuryov [24]

Answer:

I believe it is Potassium (K)

Explanation:

I did the math on a calculator and it was the closest atomic mass to potassium.

5 0
2 years ago
What is the mass of 1.82 moles of lithium carbonate?
Svetlanka [38]
You can view more details on each measurement unit: molecular weight of Lithium Carbonate or grams The molecular formula for Lithium Carbonate is Li2CO3. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Lithium Carbonate, or 73.8909 grams.
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2 years ago
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1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl
kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0
2 years ago
A solution of cough syrup contains 5.00% active ingredient by volume. If the total volume of the bottle is 11.0 mL , how many mi
Marta_Voda [28]
All you need to do is change 5% into a decimal which would be 5/100 = .05
then multiply the decimal by the total volume to get the amount of active ingredients in ml
.05 * 56ml = 2.8 ml of active ingredient.

Hope that helps!
4 0
2 years ago
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