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DENIUS [597]
3 years ago
9

Helpppppppppppppppppppppppppppppppppppppppppppppppppppp

Chemistry
2 answers:
lbvjy [14]3 years ago
4 0
Its b, not all mixtures are homogeneous 
statuscvo [17]3 years ago
3 0
The answer is C ^^ hope i helped
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6. Which of the following could be a base in a water solution?
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h20

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I did this on accident lol
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We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a
sukhopar [10]

Answer:

See Explanation

Explanation:

\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

=  236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

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ii) \frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0
3 years ago
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