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dezoksy [38]
3 years ago
7

During early research in nuclear energy, why were scientists surprised that the products from splitting a uranium nucleus had le

ss mass than the reactants ?
Chemistry
2 answers:
creativ13 [48]3 years ago
6 0
Scientists expected that the law of conservation of mass would apply to nuclear fission in terms of the masses of the subatomic particles. In reality, the mass of an atom is not equal to the sum of the masses of the subatomic particles that make it up. This is because of the energy that binds the subatomic particles. This energy has mass and when the bond is broken, the mass of the energy of the bonds is lost resulting to what we now cal, a mass defect.
Taya2010 [7]3 years ago
6 0

Answer:

3.All known changes always started and ended with the same amount of mass.

Explanation:

Here ya go!!

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How many moles of CO2 are produced from the combustion of 6.40 mol c2h8?
jarptica [38.1K]
12.8 mole of CO2 from the combustion
7 0
2 years ago
A sample of sugar (C12H22O11) contains
Oduvanchick [21]

Answer: 0.25 mol

Explanation:

Use the formula n=N/NA

n= number of mols

N =  number of particles

Nᵃ = Avogadros constant = 6.02 x

So, n=

The 10 to the power of 23 cancels out and you are left with 1.505/6.02, which is approximately 1/4. This is the same as 0.25 mol.

Hope this helped :)

5 0
3 years ago
Can you help me with this?
Nikitich [7]

Answer:

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6 0
2 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
Read 2 more answers
Boron has two naturally-ocurring isotopes. Boron-10 has an abundance of 19.8% and actual mass of 10.013 amu, and boron-11 has an
aniked [119]

Answer:

Average atomic mass = 10.812 amu

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

<u>For first isotope, Boron-10: </u>

% = 19.8 %

Mass = 10.013 amu

<u>For second isotope, Boron-11: </u>

% = 80.2 %

Mass = 11.009 amu

Thus,  

Average\ atomic\ mass=\frac{19.8}{100}\times {10.013}+\frac{80.2}{100}\times {11.009}=1.982574+8.829218=10.811792

<u>Average atomic mass = 10.812 amu</u>

6 0
3 years ago
Read 2 more answers
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