Answer:
At one atmosphere and twenty-five degrees Celsius, could you turn it into a liquid by cooling it down? Um, and the key here is that the triple point eyes that minus fifty six point six degrees Celsius and it's at five point eleven ATMs. So at one atmospheric pressure, there's no way that you're ever going to reach the liquid days. So the first part of this question is the answer The answer to the first part of a question is no. How could you instead make the liquid at twenty-five degrees Celsius? Well, the critical point is at thirty-one point one degrees Celsius. So you know, if you're twenty-five, if you increase the pressure instead, you will briefly by it, be able to form a liquid. And if you continue Teo, you know, increase the pressure eventually form a salad, so increasing the pressure is the second part. If you increase the pressure of co two thirty-seven degrees Celsius, will you ever liquefy? No. Because then, if you're above thirty-one point one degrees Celsius in temperature. You'LL never be able to actually form the liquid. Instead, you'LL only is able Teo obtain supercritical co too, which is really cool thing. You know, they used supercritical sio tu tio decaffeinated coffee without, you know, adding a solvent that you'LL be able to taste, which is really cool. But no, you can't liquefy so two above thirty-one degrees Celsius or below five-point eleven atmospheric pressures anyway, that's how I answer this question. Hope this helped :)
Answer:
V₂ = 285 mL
Explanation:
Given data:
Initial volume of bag = 250 mL
Initial temperature = 19.0°C
Final temperature = 60.0°C
Final volume = ?
Solution:
The given problem will be solved by using Charles Law,
This law stated that " The volume of given amount of gas at constant pressure and constant number of moles is directly proportional to its temperature"
Mathematical relationship:
V₁/T₁ = V₂/T₂
Now we will convert the temperature into kelvin.
Initial temperature = 19.0 + 273 = 292K
Final temperature = 60.0 + 273 = 333K
Now we will put the values in formula:
V₁/T₁ = V₂/T₂
250 mL / 292K = V₂/ 333K
0.856 mL /K = V₂/ 333K
V₂ = 0.86×333K. mL /K
V₂ = 285 mL
Answer:
4.44 g Ne
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
0.220 mol Ne
<u>Step 2: Identify Conversions</u>
Molar Mass of Ne - 20.18 g/mol
<u>Step 3: Convert</u>
<u />
= 4.4396 g Ne
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
4.4396 g Ne ≈ 4.44 g Ne
