Which of the following best describes careers that use chemistry?

What is the acknowledgment to ''Do all liquids evaporate at the same rate?'

natita [175]

'Do all liquids evaporate at the same rate
that would be false

4 0

3 years ago

Q2. 0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolpht

Vera_Pavlovna [14]

Answer:

Mass concentration (g/L) = 2.49g/L.

Explanation:

No. of moles = $\frac{mass}{molar mass}$

= $\frac{0.254}{204}$ = 0.001245 moles

Concentration of KHP (C1) in litres = n/v

= $\frac{0.001245}{0.02}$ = 0.062 mol/L

We know that:

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

where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.

Since mole ratio is 1 : 1.

1 mole of NaOH - 40g

0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g

⇒0.0498g of NaOH was used during the titration

∴Mass concentration (g/L) = 0.0498g ÷ 0.02L

= 2.49g/L.

3 0

3 years ago

5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf

mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log \frac{base}{acid}$

where, pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So, 7.4 = 7.21 + $log \frac{base}{acid}$

0.19 = $log \frac{base}{acid}$

$\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

$\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

$[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

1.55 $[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

2.55 $[H_{2}PO_{4}]$ = 1 mM

$[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

0.392 mM + $[HPO_{4}]$ = 1 mM

$[HPO_{4}]$ = (1 - 0.392) mM

$[HPO_{4}]$ = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0

3 years ago

What is the pH of 15 mL of .2 M HCl

fenix001 [56]

Easy peasy! All we need to do is plug this formula into our calculator:

-log(M)

So, we'd plug in -log(.2), which is 0.7 :)

3 0

3 years ago

Calculate the amount of energy required to melt 500 g of ice at 0oc. (δhfus=5.96 kj/mole; δhfus is the energy required to melt i

marta [7]

When you want to melt an ice, you only need the latent energy of fusion, δhfus. We use the given value, then multiply this with the given amount to determine the amount of energy. Since the energy is per mole basis, use the molar mass of ice which is 18 g/mol. The solution is as follows:

ΔH = 5.96 kJ/mol * 1 mol/18 g * 500 g
ΔH = 165.56 kJ

3 0

3 years ago