Answer:
Part A → 7.82 atm
Part B → The unknown solution had the higher concentration
Part C → 0.83 mol/L
Explanation:
Part A
Osmotic pressure (π) = M . R. T . i
NaCl → Na⁺ + Cl⁻ (i =2)
0.923 g of NaCl must be dissolved in 100 mL of solution.
0.923 g / 58.45 g/m = 0.016 moles
Molarity is mol/L → 0.016 m / 0.1L = 0.16M
π = 0.16M . 0.08206 L.atm/molK . 298K . 2 ⇒ 7.82atm
Part. B
The solvent moves toward the solution of higher concentration (to dilute it) until the two solutions have the same concentration, or until gravity overtakes the osmotic pressure, Π. If the level of the unknown solution drops when it was connected to solution in part A, we can be sure that had a higher concentration.
Part. C
π = M . R . T
20.1 atm = M . 0.08206 L.atm/mol.K . 294K
20.1 atm / (0.08206 L.atm/mol.K . 294K) = 0.83 mol/L
The first trna's anticodon loop, which will complement this mrna, is 3' GGC 5'.
<h3>What is anticodon ?</h3>
A complementary codon in messenger RNA binds to a triplet of nucleotide bases in transfer RNA that identifies the amino acid delivered during protein synthesis at a ribosome.
A codon is a trio of nucleotides, or triplet, that can be found on mRNA and codes for a certain amino acid when it is translated. A bond is created when an anticodon, which is a three-nucleotide sequence on tRNA, meets with the matching sequence on mRNA.
Each tRNA has a trio of bases known as an anti-codon. to determine the entire anti-codon sequence required to match a certain strand of mRNA. to match the complementary nucleotides in the mRNA sequence.
To learn more about anticodon from the given link:
brainly.com/question/14917337
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Answer:
2 mol H
Explanation:
For every 2 mol of NaOH, we're reacting 2 mol of H2O. In order to figure out how many mol of H are needed, it needs to be set up stochiometrically. Starting off with the given value, 1 mol of NaOH, we can then make a mol to mol ratio. For 2 mol of NaOH, we have 2 mol of H2O. For every 2 mol of H2O, we have 4 mol of H (this is because we are multiplying the coefficient by the subscript: 2 × 2). Now, we can solve for our answer.
1 mol NaOH × (2 mol H₂O / 2 mol NaOH) × (4 mol H / 2 mol H₂O)
= 2 mol H
Thus, we get 2 mol of H are needed to completely react 1 mol of NaOH.
Answer:
Volume of NaOH required = 3.61 L
Explanation:
H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:
--------(1)
where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824
--------(2)
where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000
The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.
Step 1:
Based on equation(1), at the first eq point:
moles of H2SO3 = moles of NaOH

Step 2:
For the second equivalence point setup an ICE table:

Initial 1.98 ? 0
Change -x -x x
Equil 1.98-x ?-x x
Here, ?-x =0 i.e. amount of OH- = x
Based on the Henderson buffer equation:
![pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles](https://tex.z-dn.net/?f=pH%20%3D%20pKa2%20%2B%20log%5Cfrac%7B%5BSO3%5D%5E%7B2-%7D%20%7D%7B%5BHSO3%5D%5E%7B-%7D%20%7D%20%5C%5C6.247%20%3D%207.00%20%2B%20log%5Cfrac%7Bx%7D%7B%281.98-x%29%7D%20%5C%5Cx%3D0.634%20moles)
Volume of NaOH required is:

Step 3:
Total volume of NaOH required = 3.22+0.389 =3.61 L