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Chemistry

1 answer:

qaws [65]2 years ago

7 0

Answer:

with what you didnt put a question, math equation or a attachment

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Explain how and the type of a bond that forms between sodium and chlorine in sodium chloride (NaCl)

Alexandra [31]

Explanation:

Sodium has 1 electron in its outermost shell, and chlorine has 7 electrons. It is easiest for sodium to lose its electron and form a +1 ion, and for chlorine to gain an electron, forming a -1 ion.

Now ionic bonds areIons are formed by atoms that have non-full outermost electron shells in order to become more like the noble gases in Group 8 of the Periodic Table,

Now ionic bonds areIons are formed by atoms that have non-full outermost electron shells in order to become more like the noble gases in Group 8 of the Periodic Table,Some atoms add electrons to get a full shell, thus becoming a negative ion. Other atoms subtract electrons from their outermost shell, leaving a full shell and an overall positive charge..

shell and an overall positive charge..therefore it is an ionic bond

7 0

3 years ago

Calculate the pKa of lactic acid (CH3CH(OH)COOH) given the following information. 3.005 grams of potassium lactate are added to

snow_lady [41]

Answer:

$\displaystyle \text{p} K_a \approx 3.856$

Explanation:

Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:

$\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right] & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}$

By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.

Recall the Henderson-Hasselbalch equation:

$\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}$

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for pKₐ:

$\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}$