Although the data for the experiment was not provided, we can offer a generalized answer in that when performing an experiment to achieve absolute zero temperatures, the value will never match the exact value.
<h3 /><h3>What is absolute zero?</h3>
Absolute zero is the lower limit of temperature. It is considered the coldest possible temperature that can exist. However, any attempt to reach this temperature in a controlled environment has failed, <u>scientists do not think it is possible to recreate this </u><u>temperature</u><u>. </u>
Therefore, we can confirm that the value of the absolute zero experiments did not match the accepted value. If the hypothesis was that it would be difficult or impossible to achieve, then the data would support the hypothesis, otherwise, it would fail to do so.
In summary, absolute zero is a temperature that cannot be recreated in a lab, so the value in this experiment does not match the accepted value and there is <u>no further exploration </u>to be done on this matter.
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200.0 mL =0.2000 L
Molarity = number of mole solute / volume solution(L) = 0.50 mol/0.2000 L=
= 2.5 mol/L =2.5M
Answer : 2.5 M
(1) Understanding the problem
(2) Making a plan of solution
<span>(3) Carrying out the plan </span>
The specific heat capacity of Titanium is found as 0.522 J / g°C.
<u>Explanation:</u>
We have to find the specific capacity using the formula as,
q = m × c × ΔT
Where,
q is the heat absorbed by the sample = 476 J
m is the mass of the sample = 43.56 g
c is the specific heat capacity of the sample = ?
ΔT is the temperature difference = 41.06° C - 20.13°C = 20.93° C
Now, we have to rewrite the equation to get the specific heat capacity as,
c = 
= 
= 0.522 J / g°C
So the specific heat capacity of Titanium is found as 0.522 J / g°C.
Using the equation PV = nRT
Therefore; V = nRT / P
Need moles of glucose converted to moles of the product gas (CO2).
Molecular weight calculation:
C 6 X 12.01 = 72.06
H 12 X 1.01= 12.12
O 6 X 16.00 = 96.00
sum = 180.18
25.5 g of C6H12O6 ( 1 mol C6H12O6 / 180.18 g) ( 6 mol CO2 / 1 mol C6H12O6) =
0.84915 mol CO2 gas.
Convert temp: 37 °C + 273.15 = 310.15 K
V= ((0.84915 mol)× (0.0821 L atm / mol K) (310.15 K))/0.980 atm
V = 22.0635 L
= 22.06 L CO2
