What will the final volume of a gas be if it is heated from 9 K to 117 K, and the initial volume is 85.5 L?

Chemistry

1 answer:

TiliK225 [7]3 years ago

6 0

Answer:

1111.5L

Explanation:

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Use the ideal-gas law to estimate the number of air molecules in your physics lab room, assuming all the air is N2. Assume a roo

stiv31 [10]

Answer:

4.13×10²⁷ molecules of N₂ are in the room

Explanation:

ideal gases Law → P . V = n . R . T

Pressure . volume = moles . Ideal Gases Constant . T° K

T°K = T°C + 273 → 20°C + 273 = 293K

Let's determine the volume of the room:

18 ft . 18 ft . 18ft = 5832 ft³

We convert the ft³ to L → 5832 ft³ . 28.3L / 1 ft³ = 165045.6 L

1 atm . 165045.6 L = n . 0.082 L.atm/mol.K . 293K

(1 atm . 165045.6 L) / 0.082 L.atm/mol.K . 293K = n

6869.4 moles of N₂ are in the room

If we want to find out the number of molecules we multiply the moles by NA

6869.4 mol . 6.02×10²³ = 4.13×10²⁷ molecules

4 0

2 years ago

Balance each of these equations.

Oksanka [162]

Answer:

A. 2NO + O2 -> 2NO2

B. 4Co + 3O2 -> 2Co2O3

C. 2Al + 3Cl2 -> Al2Cl6

D. 2C2H6 + 7O2 -> 4CO2 + 6H2O

E. TiCl4 + 4Na -> Ti + 4NaCl

8 0

2 years ago

A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field

maksim [4K]

Answer:

the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ )

Substituting the values into the equation, it becomes

E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ ) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²( $1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }$ ) = 7.12 × 10⁵( $1 - \frac{0.12}{0.1216}$ ) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C