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Question attached below

Lunna [17]

Answer:

D water conductivity.

3 0

4 years ago

How old is a fossil that has radioactively decayed through 4 half-lives of carbon-14?

gavmur [86]

Answer:

Carbon-14 has a half life of 5730 years. This means it takes 5730 years for half of the carbon-14 atoms to decay to carbon-12 in a given sample (eg: 100 carbon-14, one half life = 50 carbon-14, 50 carbon-12).

Explanation:

8 0

3 years ago

Read 2 more answers

Calculate the pH of a buffer solution prepared by mixing 200 mL of 0.10 M NaF and 100 mL of 0.050 M HF.

dlinn [17]

Answer:

$pH=3.74$

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one could compute the pH considering that the pKa of hydrofluoric acid, HF, is 3.14:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the concentration of the base and acid are computed by considering the mixing process with a total volume of 300 mL (0.3 L):

$n_{HF}=0.1L*0.05mol/L=0.005molHF\\\\n_{NaF}=0.2L*0.1mol/L=0.02molNaF$

$[HF]=\frac{0.005molHF}{0.3L}=0.017M$

$[NaF]=\frac{0.02molHF}{0.3L}=0.067M$

Therefore, the pH turns out:

$pH=3.14+log(\frac{0.067M}{0.017M} )\\\\pH=3.74$

Regards.

8 0

4 years ago

Use a volumetric flask to prepare 100.00 mL of 0.70M HCl solution. Do this by diluting either one of the stock hydrochloric acid

Orlov [11]

Answer:

- Take 3.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.

- Take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL

Explanation:

Hello,

In this case, given that the dilutions are preparedfrom 3.0-M and 6.0-M hydrochloric acid, we must proceed as follows:

- 3.0-M stock: when using this stock, the aliquot you must take is computed as shown below:

$V_1=\frac{M_2V_2}{M_1}=\frac{100.00mL*0.70M}{3.0M}=23.3mL$

It means that you must take 23.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.

- 6.0-M stock: when using this stock, the aliquot you must take is computed as shown below:

$V_1=\frac{M_2V_2}{M_1}=\frac{100.00mL*0.70M}{6.0M}=11.7mL$

It means that you must take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL.

Regards.

3 0

3 years ago

Calculate the solubility at 25°c of cubr in pure water and in a 0.0030m cobr2 solution. you'll find ksp data in the aleks data t

Reil [10]

Answer:

the solubility at 25°c of cubr in pure water = 0.011 g/L

The solubility at 25°c of cubr in a 0.0030m cobr2 solution = 0.00014 g/L

Explanation:

Step 1: Data given

Ksp of CuBr = 6.27 × 10^-9

Molar mass CuBr = 143.45 g/mol

Step 2: Calculate the solubility at 25°c of cubr in pure water

Ksp = [Cu+][Br-]

The initial concentration

[Cu+] = 0M

[Br-] = 0M

The concentration at the equilibrium

[Cu+] = X M

[Br-] = X M

Ksp =6.27 * 10^-9 = X * X = X²

S = 7.9 *10^-5 mol /l

7.9 * 10^-5 mol/L * 143.45 g/mol = 0.011 g/L

Step 3: Calculate the solubility at 25°c of cubr in a 0.0030m cobr2 solution.

The balanced equation:

CoBr2(aq) → Co^2+(aq) + 2 Br⁻(aq)

For 1 mol CoBr2 we'll have 1 mol Co^2+ and 2 moles Br-

For 0.0030 M CoBr2 we'll have 0.0030 M Co^2+ and 0.0060 M Br-

The initial concentration

[Cu+] = 0M

[Br-] = 0.0060 M

At the equilibrium

[Cu+] = X

[Br-] = 0.0060 + X

Ksp = 6.27 * 10^-9 = X * (0.0060+X)

6.27 *10^-9 = 0.0060 X + X²

X = 1.0 * 10^-6 mol/L

1.0 * 10^-6 mol/L * 143.45 g/mol = 0.00014 g/L

6 0

3 years ago