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Olin [163]
4 years ago
15

What makes bullet holes wider on the exit side of a piece of glass?

Chemistry
2 answers:
denpristay [2]4 years ago
6 0
The rounded shape of the bullet because as it goes through the glass it stops spinning and this make the hole wider
il63 [147K]4 years ago
6 0

Answer:

The correct answer is "The arc, or trajectory, of the bullets path".

Explanation:

During forensic examination, glass fracture analysis helps to understand the position of the attacker by analyzing how a bullet penetrated the glass. One way to understand this is by seeing where the bullet holes are narrower. Usually, bullet holes wider on the exit side of a piece of glass, which is explained by the arc, or trajectory, of the bullets path. Therefore the narrower site of the hole indicates the position at which the bullet entered the glass, and the wider site indicates the position at which the bullet exit the glass.

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9. Calculate the standard enthalpy of the 3rd reaction using the given data: A,H° = +52.96 kJ/mol AFH= -483.64 kJ/mol H2s)+I202
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Answer:

-586.56 kJ/mol is the standard enthalpy of the 3rd reaction.

Explanation:

H_2(g) +I_2(s) \rightarrow 2 HI(g) ,\Delta H^{o}_{1}= +52.96 kJ/mol...[1]

2 H_2(g) + O_2(g)\rightarrow 2 H_2O(g),\Delta H^{o}_{2}=-483.64 kJ/mol...[2]

4 HI(g)+O_2(g)\rightarrow 2 I_2(s)+2 H_2O(g) ,\Delta H^{o}_{3} =?..[3]

The unknown standard enthalpy of third reaction can be calculated by using Hess's law:

The law states that 'the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps'.

[2] - 2 × [1] = [3]

O_2+4HI\rightarrow 2H_2O(g)+2I_2(s)

\Delta H^{o}_{3}=\Delta H^{o}_{2}-2\times \Delta H^{o}_{1}

=-483.64 kJ/mol - 2\times (52.96 kJ/mol)=-586.56 kJ/mol

The standard enthalpy of the 3rd reaction is -586.56 kJ/mol.The negative sign indicates that energy is released during this reaction.

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If 0.55 g of a gas dissolved in 1.0 l of water at 20.0 kpa of pressure, how much will dissolve at 110.0 kPa of pressure?
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