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Anestetic [448]
2 years ago
9

When the u-235 nucleus is struck with a neutron, the zn-72 and sm-160 nuclei are produced, along with some neutrons. How many ne

utrons are emitted?.
Chemistry
1 answer:
Oxana [17]2 years ago
8 0

4 neutrons that are produced along with the Zn and sm. The complete final equation is:  ²³⁵U + ¹n → \;^{72}Zn+\;^{160}sm + 4 n

<h3>What atomic mass?</h3>

Atomic mass, the quantity of matter contained in an atom of an element.

1) In the left side of the transmutation equation appears:

²³⁵U + ¹n →

Deleting the atomic number (subscript to the left) because the question does not show them as it is focused on a number of neutrons.

2) The right side of the transmutation equation has:

→ \;^{72}Zn+\;^{160}sm +?

3) The total mass number of the left side is 235 + 1 = 236

4) The total mass number of Zn and sm on the right side is 160 + 72 = 232

5) Then, you are lacking 236 - 232 = 4 unit masses on the right side which are the 4 neutrons that are produced along with the Zn and sm.

The complete final equation is:

²³⁵U + ¹n → \;^{72}Zn+\;^{160}sm + 4 n

Learn more about the atomic mass here:

brainly.com/question/14250653

#SPJ1

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Answer:

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Explanation:

Isotopes can be defined as two or more forms of a chemical element that are made up of equal numbers of protons and electrons but different numbers of neutrons.

Generally, the isotopes of a chemical element have the same chemical properties because of their atomic number but different physical properties due to their atomic weight (mass number).

The two isotopes of nitrogen are nitrogen-14 and nitrogen-15.

Given the following data;

Relative abundance of N-14 = 99.63%

Atomic mass of N-14 = 14.003

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Atomic mass of N-15 = 15.000

The atomic mass is;

14.003 × (99.63/100) + 15.000 × (0.37/100)

Atomic mass = 14.003 × (0.9963) + 15.000 × (0.0037)

Atomic mass = 13.9512 + 0.0555

Atomic mass = 14.0067 amu.

<em>Therefore, the atomic mass of nitrogen is 14.0067 amu. </em>

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Explanation:

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That would be 1 mol reacting to release of ethanol,

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Now,

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Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

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