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2 years ago

3. A light bulb containing argon gas is switched on,

Chemistry

1 answer:

Readme [11.4K]2 years ago

5 0

Answer:

The assumption is quite reasonable.........

A lightbulb contains Ar gas at a temperature of 295K and at a pressure of 75kPa. The light bulb is switched on, and after 30 minutes its temperature is 418 K. What is a numerical setup for calculating the pressure of the gas inside the light bulb at 418K?

Explanation:

given constant

, and constant

, conditions that certainly obtain with a fixed volume light bulb.

And so

⋅

295

⋅

418

⋅

≅

100

⋅

Had the light bulb been sealed at normal pressure during its manufacture, what do you think might occur when it is operated?

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A gas at 750 mmhg and with a volume of 2. 00 l is allowed to change its volume at constant temperature until the pressure is 600

Gemiola [76]

Answer:

The new volume of a gas at 750 mmhg and with a volume of 2. 00 l when allowed to change its volume at constant temperature until the pressure is 600 mmhg is 2.5 Liters.

Explanation:

Boyle's law states that the pressure of a given amount of gas is inversely proportional to it's volume at constant temperature. It is written as;

P ∝ V

P V = K

P1 V1 = P2 V2

Parameters :

P1 = Initial pressure of the gas = 750 mmHg

V1 = Initial pressure of the gas = 2. 00 Liters

P2 = Final pressure of the gas = 600 mmHg

V2 = Fimal volume of the gas = ? Liters

Calculations :

V2 = P1 V1 ÷ P2

V2= 750 × 2. 00 ÷ 600

V2 = 1500 ÷ 600

V2 = 2.5 Liters.

Therefore, the new volume of the gas is 2. 5 Liters.

8 0

2 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

3 years ago

The atomic mass of helium-4 is 4.0026 amu. How many of each

umka21 [38]

Answer:

Protons: 2.

Electrons: 2.

Neutrons: 2.

Explanation:

Hello,

In this case, since an atom's atomic number is equal to the number of electrons, considering the electron configurations, taking into account that helium-4 is neither positively nor negatively charged, we can infer that the number of electrons equal the number of protons, which in this case are 2, due to the fact that is atomic number is 2.

Moreover, as helium-4's atomic number is 4 as a whole number, we compute the number of neutrons by using the shown below equation:

$Neutrons=mass\ number-atomic\ number\\\\Neutrons=4-2\\\\Neutrons=2$

Regards.

7 0

3 years ago

10. How much protein one should obtain from diet (1 gram, 3 gram,4gram)

jarptica [38.1K]

How much protein one should obtain from diet

Answer:

4 gram

7 0

3 years ago

Read 2 more answers

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$