<span>Water molecules have a lower
boiling temperature than oxygen molecules, so at room temperature they
exist as a liquid rather than a gas.
hope it helps
</span>
Answer:
2726.85 °C
Explanation:
Given data:
Initial pressure = 565 torr
Initial temperature = 27°C
Final temperature = ?
Final pressure = 5650 torr
Solution:
Initial temperature = 27°C (27+273 = 300 K)
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
T₂ = P₂T₁ /P₁
T₂ = 5650 torr × 300 K / 565 torr
T₂ = 1695000 torr. K /565 torr
T₂
= 3000 K
Kelvin to degree Celsius:
3000 K - 273.15 = 2726.85 °C
Since volume and temperature are constant, this means that pressure and <u>number of moles</u> are <u>directly </u>proportional. the sample with the largest <u>number of moles</u> will have the <u>high </u>pressure.
Since, the ideal gas equation is also called ideal gas law. So, according to ideal gas equations,
PV = nRT
- P is pressure of the sample
- T is temperature
- V is volume
- n is the number of moles
- R is universal gas constant
At constant volume and temperature the equation become ,
P ∝ nR
since, R is also constant. So, conclusion of the final equation is
P ∝ n
The number of moles and pressure of the sample is directly proportion. So, on increasing number of moles in the sample , pressure of the sample also increases.
learn about ideal gas law
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Answer:
Explanation:
If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.
Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.
However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
The attached diagrams portraying this notions is shown in the attached file below.
