2 Points

Assume that the pressure inside of the 1 liter balloon is 200,000 Pa, what is the explosive energy in the inside the balloon? Ex

AysviL [449]

Answer:

200 Joules is the explosive energy in the inside the balloon. And that is $9.523\times 10^{-5}$ 1 lb of TNT.

Explanation:

$Pressure=\frac{force}{Area}=\frac{Force\times distance}{Area \times distance}=\frac{Energy}{Volume}$

Volume of the balloon = V = 1 L = $0.001 m^3$

Pressure inside the balloon ,P= 200,000 Pa = $200,000 N/m^2$

Explosive energy in the inside the balloon be E.

E = Pressure × Volume

$E=200,000 N/m^2\times 0.001 m^3=200 Joules$

1 lb of TNT = $2.1\times 10^6 J$

200 Joules = $200\times \frac{1}{2.1\times 10^6 }$ 1 lb of TNT

= $9.523\times 10^{-5}$ 1 lb of TNT

5 0

3 years ago

What is the pressure in a 24.0-LL cylinder filled with 32.7 gg of oxygen gas at a temperature of 319 KK? Express your answer to

Blizzard [7]

Answer:

The pressure in that cylinder = 1.12atm

Explanation:

We use general gas law to calculate it. General gas law is gotten by combining Boyle's law, Charles' law and Avogadro's law. Thus

P = nRT/V

Where n = number of moles

R = the gas constant

T is the Temperature, V is the volume and P is the pressure.

Given: T = 319K, V = 24L, R = 0.0821 L.atm/K.mol

The first step is to find n using

n = mass of O2/molar mass of O2

=32.7/32

=1.0219

Now, using P =nRT/V

P = 1.0219 ×0.0821×319÷24

Therefore P = 1.12atm

8 0

3 years ago

Consider the elements: Na, Mg, Al, Si, P.

Olegator [25]

Answer:

The elements mentioned in the series are Na, Mg, Al, Si, and P. It can be seen that all these elements are located in the same period. The atomic number of the mentioned elements are Na-11, Mg-12, Al-13, Si-14, and P-15.

a) There will be an increase in the ionization energy with the increase in the element's atomic number across a period. More energy is needed to withdraw an electron from a completely occupied shell in comparison to an incompletely occupied shell.

The atomic number of Na is 11. When one electron is withdrawn from Na, it gets converted into the inert gas configuration of Ne. Thus, it will require more energy to withdraw the second electron from Na. Hence, Na exhibits the lowest second ionization energy.

b) Across the period, with an increase in the element's atomic number, the atomic radii reduces from left to right. Thus, P exhibits the smallest atomic radius.

c) The metallic nature of the elements reduces from left to right across a period in the periodic table. Thus, P is the least metallic element.

d) Diamagnetic signifies towards the element that exhibits pair electrons in its sub-shells. The electronic configuration of Mg is,

1s2 2s2 2p6 3s2

In Mg, no unpaired electrons are present, while all the remaining elements mentioned exhibit unpaired electrons in their valence shell. Thus, Mg is the diamagnetic element.

8 0

3 years ago

Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests a

fiasKO [112]

Answer:

SEPARATION SCHEME FOR CATIONS

GIVEN CATIONS : $Ag^{+} \ , Fe^{3+} , Cu^{2+}, Ni^{2+}$