<u><em>Answer:</em></u>
- The correct option is C.
- Formation of a precipitate
<u><em>Explanation:</em></u>
During a chemical reaction, new substances are formed known as a products, mostly reaction occur and their product is obtained as precipitates.
<u><em>Example</em></u>
Arylidene-2-thiobarbituric acid is obtained as precipitates when aldehyde and thiobarbituric acid react to each other.
melting of a substance
It is just indication of physical changes, like melting of ice, composition remained same as before.
boiling of a substance
It is just indication of physical changes, like boiling of water into vapors, composition remained same as before.
freezing of a substance
It is just indication of physical changes, like freezing of water into ice, composition remained same as before
The number of C atoms in 0.524 moles of C is 3.15 atoms.
The number of 
molecules in 9.87 moles is 59.43 molecules.
The moles of Fe in 1.40 x 
atoms of Fe is 0.23 x 
The moles of 
in 2.30x
molecules of is 3.81.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 × 
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
A. The number of C atoms in 0.524 mole of C:
6.02214076 × x 0.524 mole
3.155601758 atoms =3.155 atoms
B. The number of molecules in 9.87 moles of :
6.02214076 × x 9.87
59.4385293 molecules= 59.43 molecules
C. The moles of Fe in 1.40 x atoms of Fe:
1.40 x ÷ 6.02214076 ×
0.2324754694 x moles.
0.23 x moles.
D. The moles of in 2.30x molecules of :
2.30x ÷ 6.02214076 ×
3.819239854 moles=3.81 moles
Learn more about moles here:
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0.008 ÷ 51.3 = 0.0002
Sig Figs
1
0.0002
Decimals
4
0.0002
Scientific Notation
2 × 10-4
E-Notation
2e-4
Words
zero point zero zero zero two
I HOPE I HELP
Answer:i read it Explanation:
Answer:
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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%5Cleft%5B%20%5Ctext%7BKC%24_3%24H_%24_5%24O%24_3%24%7D%5Cright%5D%20%20%26%20%3D%20%5Cfrac%7B3.005%5Ctext%7B%20g%20KC%24_3%24H_%24_5%24O%24_3%24%7D%7D%7B100.%5Ctext%7B%20mL%7D%7D%20%5Ccdot%20%5Cfrac%7B1%5Ctext%7B%20mol%20KC%24_3%24H_%24_5%24O%24_3%24%7D%7D%7B128.17%20%5Ctext%7B%20g%20KC%24_3%24H_%24_5%24O%24_3%24%7D%7D%20%5Ccdot%20%5Cfrac%7B1000%5Ctext%7B%20mL%7D%7D%7B1%5Ctext%7B%20L%7D%7D%20%5C%5C%20%5C%5C%20%26%3D%200.234%5Ctext%7B%20M%7D%5Cend%7Baligned%7D)
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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%5Ctext%7BpH%7D%20%3D%20%5Ctext%7Bp%7DK_a%20%2B%20%5Clog%20%5Cfrac%7B%5Cleft%5B%5Ctext%7BBase%7D%5Cright%5D%7D%7B%5Cleft%5B%5Ctext%7BAcid%7D%5Cright%5D%7D%20%5Cend%7Baligned%7D)
[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:
In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.
