Answer:
A) 31.22
Explanation:
The reaction of sulfuric acid with NaOH is:
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O
To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.
<em>Moles H₂SO₄ (Molar mass: 98.08g/mol):</em>
0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =
3.7289x10⁻³moles H₂SO₄
And moles of NaOH that you require to neutralize the acid are:
3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =
7.4578x10⁻³ moles NaOH
Using a 0.2389M NaOH solution:
7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL
Right answer is:
<h3>A) 31.22
</h3>
Answer:
Option C= A hydrogen bond formed between a polar side chain and a hydrophobic side chain.
Explanation:
All three given options a, b and d have common mechanism to accommodate the polar amino acid.
A= A hydrogen bond forms between two polar side chains.
B= A hydrogen bond from between a polar side chain and protein back bone.
D = hydrogen bond form between polar side chains and a buried water molecules.
All these are use to accommodate the polar amino acid.
While option C is not used. which is:
A hydrogen bond formed between a polar side chain and a hydrophobic side chain.
Answer:
my mass turned into gasses and other substances besides ash
Explanation:
burning a log products smoke and ash. the smoke takes some mass and the ash takes the rest.
x= the coefficients in front of the substance in the balanced chemical equation
[H+]= the concentration of hydrogen ions
[A-]= the concentration of the other ion that broke off from the H+
[HA]= the un-disassociated acid concentration
The higher the Ka value, the greater amount of disassociation of the reactants into products. As for acids, they will break down to form H+ ions. The more the H+ ions, the stronger acidity of the solution. Thus since A has the highest Ka value, that represents the strongest acid.
You can determine the Ka value from a number of ways. If equilibrium concentrations are given of a certain acid solution, you can find the proportion of the concentration of ions to the concentration of the remaining HA molecules, using the equation above. Also, pH and KpH can be used in a number of ways. This gets more complicated and depends on the situation, and requires more advanced equations.
Hope this helped a little, its obviously not my best work
