Nucleic Acids are polymers made of nucleotideds
Answer:
1) 17.5 mL
Explanation:
Hello,
In this case, the reaction between sulfuric acid and potassium hydroxide is:

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

And in terms of concentrations and volumes:

Thus, we solve for the volume of acid:

Best regards.
Answer:
B) 0.32 %
Explanation:
Given that:

Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-

The expression for dissociation constant of acid is:
![K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BCH_3COO%7D%5E-%20%5Cright%20%5D%7D%7B%5BCH_3COOH%5D%7D)


Solving for x, we get:
<u>x = 0.00568 M</u>
Percentage ionization = 
<u>Option B is correct.</u>
2.0 L
The key to any dilution calculation is the dilution factor
The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.
In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to
DF=18.5M1.5M=12.333
So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.
The volume of the stock solution needed for this dilution will be
DF=VdilutedVstock⇒Vstock=VdilutedDF
Plug in your values to find
Vstock=25.0 L12.333=2.0 L−−−−−
The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.
So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.
IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!
In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.
Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!
So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.
Always remember
Water to concentrated acid →.NO!
Concentrated acid to water →.YES!
start the balancing by writing down how many atoms there are per element. we’ll use this as an example:
C3H8 + O2 --> H2O + CO2
C = 3 C = 1
H = 8 H = 2
O = 2 O = 3
balance the carbon first, as it is easiest to do. add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation:
C3H8 + O2 --> H2O + (3)CO2
now there are 3 carbon atoms on each side. however, when you do this, you multiply the amount of oxygen atoms you had. therefore, now, there are 6 carbon atoms in 3CO2, plus that other oxygen atom in H2O. you now have 7 O atoms instead of 3.
C = 3 C = 3
H = 8 H = 2
O = 2 O = 7
now let’s move on to the hydrogen atoms.
C3H8 + O2 --> H2O + 3CO2
you have 8 hydrogen atoms on the left side, and 2 on the right. in order to balance them, you have to multiply the right side’s hydrogen atoms by 4. 4(2) = 8.
C3H8 + O2 --> (4)H2O + 3CO2
now both hydrogen and carbon atoms are balanced. same amount on both sides. however, your oxygen atoms have changed due to the multiplying (right side). you now have 10 of them.
C = 3 C = 3
H = 8 H = 8
O = 2 O = 10
now we balance the oxygen atoms. multiply the left side of the equation’s oxygen atoms by 5. 5(2) = 10
C3H8 + (5)O2 --> 4H2O + 3CO2
the chemical equation is all balanced. basically, just multiply with numbers until it equals the same amount on both sides.
C = 3 C = 3
H = 8 H = 8
O = 10 O = 10