High pressure occur where air sinks.
Creates dry and hot conditions
The new temperature (in °C) of the gas, given the data is –148.20 °C
<h3>Data obtained from the question </h3>
- Initial temperature (T₁) = 149.05 °C = 149.05 + 273 = 422.05 K
- Initial pressure (P₁) = 349.84 KPa
- Volume = constant
- New pressure (P₂) = 103.45 KPa
- New temperature (T₂) =?
<h3>How to determine the new temperature </h3>
The new temperature of the gas can be obtained by using the combined gas equation as illustrated below:
P₁V₁ / T₁ = P₂V₂ / T₂
Since the volume is constant, we have:
P₁ / T₁ = P₂ / T₂
349.84 / 422.05 = 103.45 / T₂
Cross multiply
349.84 × T₂ = 103.45 × 422.05
Divide both side by 349.84
T₂ = (103.45 × 422.05) / 349.84
T₂ = 124.80 K
Subtract 273 from 124.80 K to express in degree celsius
T₂ = 124.80 – 273
T₂ = –148.20 °C
Learn more about gas laws:
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Molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.
The geometry around left carbon that is CH₃ is tetrahedral.
As the hybridization around left carbon is sp³ that shows its geometry should be tetrahedral and as there are 4 ligands around carbon and there is no lone pair present so the geometry is tetrahedral. So, the molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.
Answer:
On edge nuity, it is 79.
Explanation:
Please read the explanation.
Since u have no coefficients, this problem is super easy.
You just add your products, 115.5+69.91 = 185.41
Then add your reactants: 49.8+56.5= 106.3
Then subtract your reactants from your product and round to the nearest whole number.
106.3-185.41=79
It's really easy. Once u understand it, you can do any problem like this. If there is coefficients, just multiply the coefficient by the amount given to you in the problem.
