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WITCHER [35]
3 years ago
5

Atomic bombs send out a shock wave when they are detonated. This occurs because the bomb's detonation superheats the air particl

es nearby, which cause them to bump into other air particles that further the wave. Which statement can most likely be made about the shock waves of atomic bombs? They are a mechanical waves. They are an electromagnetic waves. They will act like x-rays. They will act like light rays.
Chemistry
2 answers:
ivolga24 [154]3 years ago
4 0

Answer:

A. They are mechanical waves

Explanation:

I just took the Unit test on Edge2020

brilliants [131]3 years ago
3 0

Answer:

The correct answer is - They are mechanical waves.

Explanation:

Atomic bombs after detonated they send a shock wave that makes a movement of particulate matter that leads to transfer of a high amount of energy that superheats the air medium that travels as a mechanical shock wave.

The shock wave that is sent out by a detonated atomic bomb is a longitudinal mechanical wave that transfers the energy to the neighboring particles of the air.

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Answer:

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Explanation:

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1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

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Answer:

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