1. A 4.00 L sample of air at 35 degrees C expands to 5.50 L when heated. What is the new temperature, in degrees Celsius, of the air

150.5 °C

Explanation:

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

Initial volume (V₁) = 4 L

Initial temperature (T₁) = 35 °C

Final volume (V₂) = 5.5 L

Final temperature (T₂) =?

Next, we shall convert 35 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 35 °C

Initial temperature (T₁) = 35 °C + 273

Initial temperature (T₁) = 308 K

Next, we shall determine the final (i.e the new) temperature of air. This can be obtained as follow:

Initial volume (V₁) = 4 L

Initial temperature (T₁) = 308 K

Final volume (V₂) = 5.5 L

Final temperature (T₂) =?

V₁ / T₁ = V₂ / T₂

4 / 308 = 5.5 / T₂

Cross multiply

4 × T₂ = 308 × 5.5

4 × T₂ = 1694

Divide both side by 4

T₂ = 1694 / 4

T₂ = 423.5 K

Finally, we shall convert 423.5 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T₂ = 423.5 K

T₂ = 423.5 – 273

T₂ = 150.5 °C

Thus, the new temperature of the air is 150.5 °C

gold is my favorite solid.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.