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3 years ago

How can I find the solution needed for amount of reaction?

Chemistry

1 answer:

liq [111]3 years ago

7 0

Answer:

0.48L of NaOH

Explanation:

Data obtained from the question include:

Number of mole of NaOH = 0.725mol

Molarity of NaOH = 1.50M

Volume of NaOH =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.725/1.5

Volume = 0.48L

Therefore, 0.48L of NaOH is needed for the reaction

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The pKa of the α‑carboxyl group of serine is 2.21 , and the pKa of its α‑amino group is 9.15 . Calculate the average net charge

sleet_krkn [62]

Answer:

Net charge in serine at pH equal to 8.30 is "0"

Explanation:

At pH > $pK_{a}$ , carboxyl group exists as $-CO_{2}^{-}$ (charged)

At pH < $pK_{a}$ , carboxyl group exists as -COOH (neutral)
At pH > $pK_{a}$ , amino group exists as $-NH_{2}$ (neutral)
At pH < $pK_{a}$ , amino group exists as $-NH_{3}^{+}$ (charged)
So, at pH = 8.30, both carboxyl and amino group exists in charged state.
Net charge in serine at pH equal to 8.30 is "0".
Structure of serine at pH equal to 8.30 has been shown below.

8 0

4 years ago

Diagrams, tables, and graphs are used by scientists to

Afina-wow [57]

Answer:

Since most of the data scientist collect is quantitative, data tables and charts are usually used to organize the information • Graphs are created from data tables • hope that helps love!

3 0

3 years ago

A bottle containing 415 g of cleaning solution is used to clean hospital equipment. If the cleaning solution has a specific grav

neonofarm [45]

Answer: 483 mL of the cleaning solution are used to clean hospital equipment

Explanation:

The question requires us to calculate the volume, in mL, of solution is used to clean hospital equipment, given that 415g of this solution are used and the specific gravity of the solution is 0.860.

Measurements > Density

Specific gravity is defined as the ratio between the density of a given substance to the density of a reference material, such as water:

$Specific\text{ gravity = }\frac{density\text{ of substance}}{density\text{ of reference substance}}$

The density of a substance is defined as the ratio between the mass and the volume of this substance:

$density=\frac{mass}{volume}$

Considering the reference substance as water and its density as 1.00 g/mL, we can determine the density of the substance which specific gravity is 0.860:

$0.860=\frac{density\text{ of substance}}{1.00g/mL}\rightarrow density\text{ of substance = 0.860g/mL}$

Thus, taking water as the reference substance, we can say that the density of the cleaning solution is 0.860 g/mL.

Now that we know the density of the cleaning solution (0.860 g/mL) and the mass of solution that is used to clean hospital equipment (415g), we can calculate the volume of solution that is used to clean the equipment:

$\begin{gathered} density=\frac{mass}{volume}\rightarrow volume=\frac{mass}{density} \\ \\ volume=\frac{415g}{0.860g/mL}=483mL \end{gathered}$

Therefore, 483 mL of the cleaning solution are used to clean hospital equipment.

6 0

1 year ago

A stack of 15 pennies is immersed into a 100 mL graduate initially containing 20.6 mL of water. The volume in the graduate cylin

Gekata [30.6K]

Answer: 5.4

Explanation: The Law of Impenetrability says that two objects can't occupy the same space at the same time; therefore, the pennies and the water can't occupy the same space at the same time. Without the pennies the water level read 20.6 mL, dropping in the pennies gives a level of 26.0.

5 0

3 years ago

What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?

xz_007 [3.2K]

Answer:

B) 0.32 %

Explanation:

Given that:

$K_{a}=1.8\times 10^{-5}$

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

$\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}$

The expression for dissociation constant of acid is:

$K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}$

$1.8\times 10^{-5}=\frac{x^2}{1.8-x}$

$1.8\left(1.8-x\right)=100000x^2$

Solving for x, we get:

Percentage ionization = $\frac{0.00568}{1.8}\times 100=0.32 \%$

<u>Option B is correct.</u>

8 0

4 years ago