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3 years ago

8

Nuclear power plants produce energy using fission. One common fuel, uranium-235, produces energy through the fission reaction 23

5 92U+10n→fission fragments+neutrons+3.20×10−11 J/atom 92235U+01n→fission fragments+neutrons+3.20×10−11 J/atom What mass of uranium-235 is needed to produce the same amount of energy as the fusion reaction in Part A? Express your answer to three significant figures and include the appropriate units.

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

mass of U-235 = 15.9 g (3 sig. figures)

Explanation:

1 atom can produce -------------------------> 3.20 x 10^-11 J energy

x atoms can produce ----------------------> 1.30 x 10^12 J energy

x = 1.30 x 10^12 / 3.20 x 10^-11

x = 4.06 x 10^22 atoms

1 mol ----------------------> 6.023 x 10^23 atoms

y mol ----------------------> 4.06 x 10^22 atoms

y = 0.0675 moles

mass of U-235 = 0.0675 x 235 = 15.8625

mass of U-235 = 15.9 g (3 sig. figures)

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What is the mass of silver (107.87 g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate?Cu

vladimir2022 [97]

Answer: 323.61 g of $Ag$ will be produced

Explanation:

The given balanced chemical reaction is :

$Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag$

According to stoichiometry :

2 moles of $AgNO_3$ require 1 mole of $Cu$

Thus 3.00 moles of $AgNO_3$ will require= $\frac{1}{2}\times 3.00=1.50moles$ of $Cu$

Thus $AgNO_3$ is the limiting reagent as it limits the formation of product.

As 2 moles of $AgNO_3$ give = 2 moles of $Ag$

Thus 3.00 moles of $AgNO_3$ give = $\frac{2}{2}\times 3.00=3.00moles$ of $Ag$

Mass of $Ag=moles\times {\text {Molar mass}}=3.00moles\times 107.87g/mol=323.61g$

Thus 323.61 g of $Ag$ will be produced from the given moles of both reactants.

3 0

3 years ago

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0

3 years ago

Aides in the burning of materials. Needed for humans to breathe.

Veseljchak [2.6K]

Answer:

Oxygen.

Explanation:

Oxygen is a chemical element that aides in the burning of materials and it is needed for humans to breathe.

Cellular respiration can be defined as a series of metabolic reactions that typically occur in cells so as to produce energy in the form of adenosine triphosphate (ATP). During cellular respiration, high energy intermediates are created that can then be oxidized to make adenosine triphosphate (ATP). Therefore, the intermediary products are produced at the glycolysis and citric acid cycle stage through the breathing of oxygen used to obtain energy from the food ingested.

Hence, all cells in living organisms require oxygen and glucose to release energy.

3 0

3 years ago

Does covalency fail in explaining the formation of BeCl2 and PCl5? Explain

Stolb23 [73]

Answer:

Yes , covalency fails to explain the formation of BeCl₂ and PCl₅.

Explanation:

covalent bonds are formed between two atoms by sharing of outermost electrons in order to attain octet for attaining stability . To octet means to attain 8 electrons in their outermost orbit .

BeCl₂ :

Be has two electrons in its outermost orbit and each of Cl has 7 electrons .

Each of Cl shares one electron each with each of two electrons of Be . Cl atoms attain 8 electrons as octet . But Be attains only 4 electrons ( 2 of Be and 2 from Cl atoms ) . Hence octet of Be is not attained . So covalency fails in the formation of BeCl₂ .

PCl₅ :

P has 5 electrons in the outermost orbit . Each of these electrons shares with one electron from each of Cl atoms . Thus Cl attains octel state ( 8 electrons ) but P attains 10 electrons ( 5 + 5 ) , 5 of P and 5 from 5 Cl atoms .

Hence Octet of P is not attained . Hence covalency fails in the formation of PCl₅ .

8 0

3 years ago

A change in which property of light will have no effect on whether or not the photoelectric effect occurs

AURORKA [14]

The intensity<span> of the light has no connection with the photoelectric effect.</span>

<span>That's what was so baffling about it before the particle nature of light </span>
<span>was suspected ... a match with a blue flame might stimulate the </span>
<span>photoelectric effect, but a high-power red searchlight couldn't do it.</span>

3 0

3 years ago