Does the amount of Borax added change the slime?

Consider the electrolysis of molten barium chloride, BaCl2. (a) Write the half-reactions. (b) How many grams of barium metal can

Zielflug [23.3K]

Answer: a) Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) 0.640 grams of Ba will be deposited.

Explanation: a) The problem is based on Faraday law of electrolysis. Molten barium chloride has $Ba^+^2$ ion and $Cl^-$ ion. Barium ion is reduced and chloride is oxidized. Reduction takes place at cathode and oxidation at anode. So, the half reactions will be:

Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) The question asks, how many grams of barium metal can be produced by supplying 0.50 ampere for 30 minutes.

From the Cathode half-reaction, 1 mole of Ba is deposited by 2 moles of electrons and we know that 1 mole of electron carries one Faraday that is 96485 Coulomb.

Coulombs for 2 moles of electrons will be = 2*96485 C = 192970 C

So, we can say that, 192970 C will deposit 1 mole of Ba metal.

Total available coulombs can be calculated using the formula:

q=i*t

where, q is electric charge in coulomb, i is current in ampere and t is time in seconds.

$q=q=0.50A*30min(\frac{60sec}{1min})$

q = 900 C (note: 1 C = 1 A*sec)

Let's calculate how many moles of Ba will get deposited by 900 C.

$900C(\frac{1molBa}{192970C})$

= 0.00466 mole Ba

Convert the moles of Ba to grams and for this we multiply by molar mass of Ba which is 137.33 gram per mol.

$0.00466molBa(\frac{137.33g}{1mol})$

= 0.640 g Ba

So, 0.50 A for 30 minutes will deposit 0.640 grams of Ba metal.

6 0

2 years ago

Question 3

Dmitrij [34]

Answer:

snow and very low temperatures are predicted for tomorrow.

Explanation:

snow and very low temperatures are associated with average weather condition of a place thus making it a statement about climate.

4 0

3 years ago

The electron configuration of an element is 1s^ 2 2s^ 2 2p^ 2 How many valence electrons does the clement have? (1 point )

Monica [59]

Answer:

:)

Explanation:

8 0

2 years ago

What is the mass of HF produced by three reaction of 3.0 10 to the 23 molecules of H2 with excess F2

JulijaS [17]

Answer:

It is 20. g HF

Explanation:

H2 + F2 ==> 2HF ... balanced equation

Since the question is asking us to find the mass of product formed, we will want to first convert the molecules of H2 into moles of H2 (we could do this at the end of the calculations, but it's just as easy to do it now).

moles of H2 present (using Avogadro's number):

3.0x1023 molecules H2 x 1 mole H2/6.02x1023 molecules = 0.498 moles H2

From the balanced equation, we see that 1 mole H2 produces 2 moles HF. Therefore, we can now find the theoretical mass of HF produced from 0.498 moles H2:

0.498 moles H2 x 2 moles HF/1 mol H2 = 0.996 moles HF formed.

The molar mass of HF = 20.01 g/mole, thus...

0.996 moles HF x 20.01 g/mole = 19.93 g HF = 20. g HF formed (to 2 significant figures)

6 0

2 years ago

Carbon disulfide is prepared industrially by reacting carbon with sulfur dioxide according to the above equation. If 70.8 g of c

Rus_ich [418]

Answer:

1.18 moles of CS₂ are produced by the reaction.

Explanation:

We present the reaction:

5C + 2SO₂ → CS₂ + 4CO

5 moles of carbon react to 2 moles of sulfur dioxide in order to produce 1 mol of carbon disulfide and 4 moles of carbon monoxide.

As we do not have data from the SO₂, we assume this as the excess reagent. We convert the mass of carbon to moles:

70.8 g / 12 g/mol = 5.9 moles

Ratio is 5:1, so 5 moles of carbon react to produce 1 mol of CS₂

Then, 5.9 moles will produce (5.9 . 1) / 5 = 1.18 moles

8 0

3 years ago