The sample has a new pressure of 274kPa. If at 105 kPa and 275K, a 220 mL sample of helium gas is contained in a cylinder with a moving piston. The sample is pushed till it has a 95.0 mL volume and 310K .
The macroscopic characteristics of ideal gases are related by the ideal gas law (PV = nRT). A gas is considered to be perfect if its particles (a) do not interact with one another and (b) occupy no space (have no volume). Where P= pressure V= volume and T = temperature.
From ideal gas equation
P₁V₁/T₁ =P₂V₂/T₂
105×220÷275 = P₂ ×95÷310
P₂= (105×220×310)÷(275×95)
P2= 7161000/26125
P2 = 274.105 kPa
Hence, the new pressure of helium gas is 274kPa
To know more about Ideas gas equation
brainly.com/question/28837405
#SPJ1
Answer:
The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.
Explanation:
The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.
There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.
All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".
In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.
The other options are correctly written.
<span>The correct answer is 'freezing point depression'. Colligative properties depend on the concentration of molecules of a solute. Examples of other colligative properties are boiling point elevation or vapour pressure lowering. The salt causes ice on the side walk to melt because it lowers the freezing point. </span>
Answer:
T2 = 550K
Explanation:
From Charles law;
V1/T1 = V2/T2
Where;
V1 is initial volume
V2 is final volume
T1 is initial temperature
T2 is final temperature
We are given;
V1 = 20 mL
V2 = 55 mL
T1 = 200 K
Thus from V1/T1 = V2/T2, making T2 the subject;
T2 = (V2 × T1)/V1
T2 = (55 × 200)/20
T2 = 550K
