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2 years ago

The 225 g FeCl2 is about 1.8 moles.

Chemistry

1 answer:

Lyrx [107]2 years ago

5 0

Answer:

$4 \frac{mol}{l}$

Explanation:

From the question, we have been asked to find the molarity of FeCl2 having a volume of 450 mL,

We have been provided with 225 g which is proportional to 1.8 moles.

We know that molarity of any solution should be in mol/L.

1 mole contained in 1 L means it has a molarity of 1 mol/L

Let's convert 450 mL to Litres which is,

$\frac{450}{1000} litres$

= 0.450 L

Thus,

1 mole is contained in 1L

x moles are contained in 0.450 L

Hence,

x mole/molarity = {1 mole x 1 L}/{0.450 L}

= 4 mol/L

Therefore 4 mol/L is the molar concentration.

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The table shows the amount of radioactive element remaining in a sample over a period of time.

Assoli18 [71]

It would take 147 hours for 320 g of the sample to decay to 2.5 grams from the information provided.

Radioactivity refers to the decay of a nucleus leading to the spontaneous emission of radiation. The half life of a radioactive nucleus refers to the time required for the nucleus to decay to half of its initial amount.

Looking at the table, we can see that the initial mass of radioactive material present is 186 grams, within 21 hours, the radioactive substance decayed to half of its initial mass (93 g). Hence, the half life is 21 hours.

Using the formula;

k = 0.693/t1/2

k = 0.693/21 hours = 0.033 hr-1

Using;

N=Noe^-kt

N = mass of radioactive sample at time t

No = mass of radioactive sample initially present

k = decay constant

t = time taken

Substituting values;

2.5/320= e^- 0.033 t

0.0078 = e^- 0.033 t

ln (0.0078) = 0.033 t

t = ln (0.0078)/-0.033

t = 147 hours

Learn more: brainly.com/question/6111443

7 0

2 years ago

Draw an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. (Assume standard conditions).

babunello [35]

Answer:

-1.05 V

Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

Oxidation half equation;

Sn(s) ------> Sn^2+(aq) + 2e

Reduction half equation:

Mn^2+(aq) + 2e ----> Mn(s)

Cell voltage= E°cathode - E°anode

E°cathode= -1.19V

E°anode= -0.14 V

Cell voltage= -1.19 V - (-0.14V)

Cell voltage= -1.05 V