. How do metals and non-metals

K = °C + 273 A 4.1 L sample of gas is held at 25 °C. If the gas expands to 6.8 L, what is the final temperature?

gregori [183]

Answer:

221 °C

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 4.1 L

Initial temperature (T₁) = 25 °C

= 25 °C + 273

= 298 K

Final volume (V₂) = 6.8 L

Final temperature (T₂) =?

The final temperature of the gas can be obtained as follow:

V₁ / T₁ = V₂ / T₂

4.1 / 298 = 6.8 / T₂

Cross multiply

4.1 × T₂ = 298 × 6.8

4.1 × T₂ = 2026.4

Divide both side by 4.1

T₂ = 2026.4 / 4.1

T₂ ≈ 494 K

Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:

°C = K – 273

K = 494

°C = 494 – 273

°C = 221 °C

Thus the final temperature of the gas is 221 °C

6 0

3 years ago

Whats the formula for mass

s2008m [1.1K]

Answer:

The unit of molar mass is kg/mol. Where, m = mass, v = velocity.

Hope that helps :)

5 0

2 years ago

draw the lewis structure for CO2, H2CO3, HCO3-, and CO3 2-.Rank these in order of increasing attraction to water molecules. Expl

gavmur [86]

Answer:

The structures are attached in file.

Hydrogen bonding and intermolecular forces is the reason for ranks allotted.

Explanation:

In determining Lewis structure, we calculate the overall number of valence electrons available for bonding. Making carbon (the least electronegative atom) the central atom in the structure, we allocate valence electrons until each atom has achieved stability.

In order of decreasing affinity to water molecules:

$CO_{3}^{2-} > HCO_{3} ^{2-} > H_{2} CO_{3}$

This is due to the fact that the $CO_{3}^{2-}$ will accept protons more readily than the bicarbonate ion, $HCO_{3} ^{2-}$ . Carbonic acid, $H_{2} CO_{3}$ will not accept any more protons, hence it is the least attractive to water molecule, even though soluble.