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Alex
3 years ago
10

in terms of amounts of energy and harmful radioactive waste produced , which of the following is the most accurate comparison of

fission and fusion
Chemistry
1 answer:
Eddi Din [679]3 years ago
3 0

Explanation:

Nuclear fusion produces more energy with less radioactive waste compared to nuclear fission process.

  • In nuclear fission, a radioactive decay occurs in which a heavy nucleus spontaneously disintegrates into two lighter nuclei and several neutrons with the release of energy.
  • Nuclear fusion occurs when small atomic nuclei combines into larger ones with the release of a large amount of energy.
  • Instability of a nuclide is enough to start up a nuclear fission reaction. In some instances, a heavy nuclide is bombarded with another nuclide.
  • In nuclear fusion, the activation energy required to start up is very great.

Nuclear wastes emanating from nuclear fusion process is least compared to that of nuclear fission.

Nuclear fusion produces more energy  compared to nuclear fission process.

Learn more:

Nuclear reaction brainly.com/question/10094982

#learnwithBrainly

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1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

Molarity=\frac{Moles}{Volume (L)}

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

3 0
4 years ago
How was the structure of the atom discovered
Gennadij [26K]

Answer:

read down below

Explanation:

Building on the Curies' work, the British physicist Ernest Rutherford (1871–1937) performed decisive experiments that led to the modern view of the structure of the atom. ... Because it was the first kind of radiation to be discovered, Rutherford called these substances α particles.

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DUE TOMORROW!!! 15 POINTS
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Answer:

A. write balanced chemical equation (including states), for this process.

Explanation:

Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.  

Hydrocarbon combustions always involve  

[some hydrocarbon] + oxygen --> carbon dioxide + steam.  

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

Balance carbon, six on each side:  

C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

Balance hydrogen, six on each side:  

C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

Now, we have fifteen oxygens on the right and O2 on the left.  

Two ways to deal with that. We can use a fraction:  

C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)

Or, if you prefer to have whole number coefficients, double everything  

to get rid of the fraction:  

2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

With the SATP states thrown in...  

C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)

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bezimeni [28]
Mitochondria need the most oxygen
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Xelga [282]

Answer:

1. B.10 mol/l

2. D.

3.A.

Explanation:

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