2.2311 moles of gas are there in a 50. 0 l container at 22. 0 °c and 825 torrs.
<h3>What is an ideal gas?</h3>
An Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Assuming the gas is ideal, we can solve this problem by using the following equation:
PV = nRT
Where:
P = 825 torr ⇒ 825 / 760 = 1.08 atm
V = 50 L
n = ?
R = 0.082 atm·L·mol⁻¹·K⁻¹
T = 22 °C ⇒ 22 + 273.16 = 295.16 K
We input the data:
1.08 atm x 50 L = n x 0.082 atm·L·mol⁻¹·K⁻¹ x 295.16 K
And solve for n:
24.20312
n = 2.2311 mol
Hence, 2.2311 moles of gas are there in a 50. 0 L container at 22. 0 °c and 825 torrs.
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Answer:
To understand the utility in sequence comparison and in the search for proteins that have a common evolutionary origin, you need to be clear about some concepts about how to evolve proteins. The idea that is accepted is that throughout the evolution some species are giving rise to new ones. Behind this is the genetic variation of organisms, that is, the evolution of genomes and their genes, as well as the proteins encoded by them.
Explanation:
Three ways can be distinguished by which genes evolve, and by proteins: mutation, duplication and shuffling of domains. When differences between homologous protein sequences are observed, these differences change to do with the way of life of the organism, an example of this, bacteria that live in hot springs at very high temperatures have proteins with a very high denaturation temperature, and these proteins are usually richer in cysteines. On the other hand, the fact that in positions of the sequences they remain unchanged (preserved positions), means that these have a special importance for the maintenance of the structure or function of the protein and its modification has not been tolerated throughout of evolution
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
Push the pushbutton to the first stop before inserting the tip into the solution, and then put the tip in the solution tube.
