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kvv77 [185]
2 years ago
3

A carbon-12 atom has a mass defect of 0.09564 amu. Which setup is used to calculate nuclear binding energy?

Chemistry
2 answers:
EastWind [94]2 years ago
6 0
A carbon-12 atom has 6 protons, it differs from the regular carbon in the number of neutrons, it ha 6 neutrons.
The binding energy is the energy needed to put in to split the nucleus into individual protons and neutrons. To get the binding energy we add the masses of individual protons, neutrons and electrons, then subtract the mass of the atom, and convert that mass difference to energy. In other words we convert the mass defect into energy.  1 amu = 1.660540 × 10^-27 kg, and the mass defect is 0.09564 amu
Therefore the correct answer is a;
<span>0.09564 amu x (1.6606 x 10–27 kg)/amu x (3.0 x 10^8)</span>²

dsp732 years ago
6 0

a. <u>0.09564 amu x  (1.6606 x  10–27 kg)/amu x (3.0 x  108)2  </u>

b. 0.09564 amu x  1 amu/(1.6606 x  10–27 kg) x  (3.0 x 108)2

c. 0.09564 amu x  (3.0 x 108)2

d. 0.09564 amu x  (1.6606 x  10–27 kg)/amu x  (3.0 x  108)  

Answer: first option (a) <u>0.09564 amu x  (1.6606 x  10–27 kg)/amu x (3.0 x  108)2  </u>

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What is the halflife of a radioisotope if a 20-g sample becomes 10g after 16 hours
ohaa [14]

Answer:

T½ = 16hours

Explanation:

Final mass (N) = 10g

Initial mass (No) = 20g

Time (t) = 16hours

T½ = ?

T½ = In2 / λ

But λ = ?

In(N/No) = -λt

In(10/20) = -(λ * 16)

In(0.5) = -16λ

-0.693 = -16λ

λ = 0.693 / 16

λ = 0.0433

Note : λ is known as the disintegration constant

T½ = In2 / λ

T½ = 0.693 / 0.0433

T½ = 16hours

The half-life of the sample is 16hours

5 0
2 years ago
How many grams of antifreeze C2H4(OH)2 would be required per 500 g of water to prevent the water from freezing at a temperature
Wewaii [24]

Answer:

333.7 g.

Explanation:

  • The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>

Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).

Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>

molaity (m) is the no. of moles of solute per kg of the solvent.

∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))

∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.

∵ no. of moles = mass/molar mass.

<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>

5 0
2 years ago
a 16.0g sample of iron was heated from 0 degrees Celsius o 35.0 degrees Celsius.the iron absorbed 246.4J of energy as heat. what
frosja888 [35]
Specific heat means the amount heat needed when unit mass of a substrate increase one degree of temperature. So the specific heat = the heat absorbed/(the mass of the substrate * change in temperature) = 264.4/(16*35)=0.472 J/(g*℃)
5 0
2 years ago
Read 2 more answers
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
A sample of gas contains 0.1800 mol of CO(g) and 0.1800 mol of NO(g) and occupies a volume of 23.2 L. The following reaction tak
baherus [9]

Answer:

The volume of the sample is 17.4L

Explanation:

The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:

0.1800mol + 0.1800mol reactants =

0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.

Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:

V1n2 = V2n1

<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>

Replacing:

V1 = 23.2L

n2 = 0.2700 moles

V2 = ??

n1 = 0.3600 moles

23.2L*0.2700mol = V2*0.3600moles

17.4L = V2

<h3>The volume of the sample is 17.4L</h3>
8 0
1 year ago
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