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3 years ago

How do I prepare for my general chemistry labs? How do I get good grades on labs?

2 answers:

8 0

Well you should study how different chemicals work, and make flashcards to carry around or make notes during class to study for later

Reika [66]3 years ago

6 0

You must make sure that you pay attention in class, follow directions, do all homework and assignments to the best of your abilities, listen to the teacher, ask for help when it is needed, now what is going on in that class, do not daze of and focus, understand the subject, and try your hardest. You can get good grades by following all directions, completing it, putting lots of effort and detail into it, asking for help,and doing it the best that you possibly can. Good Luck I'm sure you can do it!!

You might be interested in

What is the ratio of lactic acid (Ka = 1.37x^10-4) to lactate in a solution with pH =4.29

hram777 [196]

Henderson–Hasselbalch equation is given as,

pH = pKa + log [A⁻] / [HA] -------- (1)

Solution:

Convert Ka into pKa,

pKa = -log Ka

pKa = -log 1.37 × 10⁻⁴

pKa = 3.863

Putting value of pKa and pH in eq.1,

4.29 = 3.863 + log [lactate] / [lactic acid]

Or,

log [lactate] / [lactic acid] = 4.29 - 3.863

log [lactate] / [lactic acid] = 0.427

Taking Anti log,

[lactate] / [lactic acid] = 2.673

Result:

2.673 M lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.

6 0

3 years ago

If 4.12 l of a 0.850 m-h3po4 solution is be diluted to a volume of 10.00 l, what is the concentration of the resulting solution?

statuscvo [17]

The process in which the concentration of the solution is lessened by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.

Formula is given by:

$C_{1}V_{1}=C_{2}V_{2}$ (1)

where,

$C_{1}$ is the initial concentration

$V_{1}$ is the initial volume

$C_{2}$ is the final concentration

$V_{2}$ is the final volume

Now,

$C_{1}$ = 0.850 M

$V_{1}$ = 4.12 L

$C_{2}$ =?

$V_{2}$ = 10.00 L

Substitute the give values in formula (1),

$0.850 M\times 4.12L=C_{2}\times 10.00 L$

$C_{2} =\frac{0.850 M\times 4.12L}{10.00 L}$

= $0.3502 M$

Thus, the final concentration of the $H_{3}PO_{4}$ solution = $0.3502 M$

5 0

3 years ago

2KCIO3 -2KCL + 302

Firlakuza [10]

Answer:

a. 6 moles of O₂

b. 4.06×10²⁴ molecules of O₂

c. 3.04 g of KCl

Explanation:

Reaction of decomposition is:

2KClO₃ → 2KCl + 3O₂

a. See stoichiometry value

2 moles of potassium chlorate can decompose to 2 moles of potassium chloride and 3 moles of oxygen. Ratio is 2:3

If 2 moles of KClO₃ can decompose to 3 moles of O₂

Then 4 moles, may decompose to (4 . 3)/2 = 6 moles of O₂

b. In this case, the stoichiometry is the same.

Per 2 moles of KClO₃, I produce 2 moles of KCl

Then, 4.5 moles of KCl, were produced by 4.5 moles of KClO₃

We apply, the last relation:

(4.5 . 3) /2 = 6.75 moles of O₂ are also produced.

How many molecules are in 6.75 moles?

6.75 mol . 6.02×10²³ molecules/mol = 4.06×10²⁴ molecules of O₂

c. First of all, we convert the mass to moles:

5g . 1mol /122.55g = 0.0408 moles of salt

As ratio is 2:2, 0.0408 moles of salt, decompose to 0.0408 moles of KCl

We convert the moles to mass: 0.0408 mol . 74.55g /mol = 3.04g

5 0

2 years ago

50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the

scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

$[H_3}^+NC_2H_5CO^-_2]$ = 0.08 M

Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{8}{150}$

$[H^+_3NC_2H_5CO_2H]$ = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

$H^+_3NC_2H_5CO_2H$ ⇄ $H^+_3NC_2H_5CO^-_2$ $+$ $H^+$

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So $[H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M$

pH = $-log[H^+]$

pH = $-log[0.0134]$

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0

3 years ago

vaporized at 100°C and 1 atmosphere pressure. Assuming ideal gas 1 g mole of water is behavior calculate the work done and compa

Vesna [10]

Answer:

q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.

Explanation:

1. Heat absorbed

q = nΔH = 1 mol × (40.57 kJ/1 mol) = 40.57 kJ

2. Change in volume

V(water) = 0.018 L

pV = nRT

1 atm × V = 1 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 373.15 K

V = 30.62 L

ΔV = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L

3. Work done

w = -pΔV = - 1 atm × 30.60 L = -30.60 L·atm

w = -30.60 L·atm × (101.325 J/1 L·atm) = -3100 J = -3.10 kJ

4. Why the difference?

Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.

The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.

7 0

3 years ago