Answer:
36.2 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 8.6 atm
- Initial temperature of the gas (T₁): 38°C
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final temperature of the gas (T₂): ?
Step 2: Convert T₁ to Kelvin
We will use the following expression.
K = °C +273.15
K = 38 °C +273.15 = 311 K
Step 3: Calculate T₂
We will use Gay Lussac's law.
P₁/T₁ = P₂/T₂
T₂ = P₂ × T₁/P₁
T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K
Answer:
Yes
Explanation:
There is a difference between the homogeneous mixture of the hydrogen and the oxygen in a 2:1 ratio and the sample of the water vapor.
In the homogeneous mixture of the hydrogen and the oxygen which are present in the ratio, 2:1 , the elements are not chemically combined. They are explosive also as both shows their specific properties. They can be separated by physical means (Condensation, diffusion).
On the other hand, in water vapor, the two elements are chemically bonded in a specific mixture which cannot be separated via physical means. Water has its unique properties and they can be separated by chemical means only.
Answer:
Its a physical
theres no chemical properties involved.
Explanation:
Brainliest please
Answer:
See explanation
Explanation:
We can describe electrons using four sets of quantum numbers;
principal quantum number (n)
orbital quantum number (l)
magnetic quantum number (ml)
spin quantum number (ms)
Since no two electrons in an atom can have the same value for all four quantum numbers according to Pauli exclusion theory, for the orbitals given one possible value for each quantum number is shown below;
For 1s-
n = 1, l= 0, ml = 0, ms= 1/2
For 2s-
n= 2, l =0, ml=0, ms=1/2
For 1s and 2s orbitals, there is only one possible value for ml which is zero.
