All categories

3 years ago

What is the difference between an airburst nuclear bomb and a surface bomb?

Chemistry

2 answers:

Alona [7]3 years ago

8 0

Answer:

they are all explosion but different materials

Explanation:

KiRa [710]3 years ago

4 0

Nuclear explosions are generally classified as air bursts, surface bursts, subsurface bursts, or high altitude bursts. ... Air Bursts. An air burst is an explosion in which a weapon is detonated in air at an altitude below 30 km but at sufficient height that the fireball does not contact the surface of the earth.

You might be interested in

What is thr valency of scadium

Hoochie [10]

scandium is a transition metal and shows a valency of 3.

4 0

3 years ago

Who is all bor.ed bc im bor.ed who wan.na talk UwU

emmainna [20.7K]

Answer:

me ig

Explanation:

4 0

3 years ago

Read 2 more answers

Salt is dissolved in a flask of tap water. Distilling the mixture causes the salt to separate from the water. Which type of ener

Alik [6]

Heat energy is required.
In distillation, the solution is first heated, where heat energy is required, such as using a bunsen burner.
When the solution is heated, the water may reach its boiling point and evaporate. However, salt does not. When water molecules evaporates, it travels through a condenser that cools it down into liquid again. Therefore we get pure water. Salt is also obtained in the original beaker.
Therefore to first start this process, heat energy is required.

6 0

3 years ago

Read 2 more answers

What is the volume occupied by 2.50 moles of oxygen gas at STP

oee [108]

1 mole of any gas under STP has volume 22.4 L
So 2.50 moles of any gas ( including oxygen)
2.50 mol *(22.4L/1 mol)=56.0 L

8 0

3 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago