To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.
The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L
Rearranging for T1:
T1 = (V1 × T2) / V2
Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C
Answer:
Ir(NO2)3
Explanation:
The molar mass is 330.2335, in case that's also required.
The answer is C. The specific amount of energy released when excited electrons fall back to the ground state produces an emission spectrum. That energy is emitted as photons with precise wavelengths corresponding to the energy differences between levels. Because each element yields a characteristic set of wavelengths, the emission spectrum can be used to identify the element in the sample.
Response:
4.5 m³
Resolution:
The statement indicates the presence of two blocks on a lid of a container with a volume of 9 m³. The lid's weight is equal to that of the two blocks. Thus, there were initially four blocks (or 4 atm pressure) acting on a volume of 9 m³.
After adding four additional blocks on the lid, the pressure rises from 4 atm to 8 atm (2 atm from the lid, 2 atm from the original blocks, and 4 atm from the new blocks).
Hence, The data established is,
P₁ = 4 atm
V₁ = 9 m³
P₂ = 8 atm
V₂ =?
Using Boyle's Law,
P₁ V₁ = P₂ V₂
Resolving for V₂,
V₂ = P₁ V₁ / P₂
Substituting values yields:
V₂ = (4 atm × 9 m³) ÷ 8 atm
V₂ = 4.5 m³