Why can’t you do electrolysis with AQUEOUS lead bromide?

Convert the following into balanced equations:

enyata [817]

Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃

We must first convert from a word equation to a symbol equation:to a symbol equation:

Lead (II) Nitrate + Potassium Iodide → Lead (II) Iodide + Potassium Nitrate

The lead (II) ion is represented as Pb²⁺ , whilst the nitrate ion is NO⁻₃

To balance the charges, we require two nitrate ions per lead (II) ion, and so lead (II) nitrate is Pb(NO₃)₂

The potassium ion is K ⁺ and the iodide ion is I ⁻

The two charges balance in 1:1 ratio, giving a formula of KNO₃

The symbol equation is as follows:

Pb(NO₃)₂ + KI →PbI₂ + KNO ₃

The most obvious change we must make, when balancing this equation, is to increase the number of nitrate ions on the right hand side of the equation. We can to this by placing a coefficient of 2 before the potassium nitrate:

Pb(NO₃)₂ +KI →PbI₂ +2KNO₃

In doing this we have upset the balance of potassium ions on each side of the equation.

Again, we can fix this: we must simply place another coefficient of 2, this time before the potassium iodide:

Pb(NO₃)₂ +2KI → PbI₂ +2KNO ₃

Concluding :

Checking over the equations once more, you will notice that we initially had 1 iodide ion on the right hand side, but 2 on the left. However, we already dealt with this in balancing out potassium ions. Now, our equation is balanced.

And that's it! One last thing to add is that you may have noticed the irregularity in iodide ions rather than nitrate ions. In this case, you would arrived at the same answer simply by working backwards.

Learn more about balanced equation :

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3 0

2 years ago

A 420 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 7 mL of 1.00 M KOH. What is the pH following this addition

Art [367]

Answer: The pH of the resulting solution will be 3.60

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$ ......(1)

We are given:

Molarity of formic acid = 0.100 M

Molarity of potassium formate = 0.100 M

Volume of solution = 420 mL = 0.420 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$\text{Moles of formic acid}=(0.100mol/L\times 0.420L)=0.0420mol$

$\text{Moles of potassium formate}=(0.100mol/L\times 0.420L)=0.042mol$

Molarity of KOH = 1.00 M

Volume of solution = 7 mL = 0.007 L

Putting values in equation 1, we get:

$\text{Moles of KOH}=(1mol/L\times 0.007L)=0.007mol$

The chemical equation for the reaction of formic acid and KOH follows:

$HCOOH+KOH\rightleftharpoons HCOOK+H_2O$

I: 0.042 0.007 0.042

C: -0.007 -0.007 +0.007

E: 0.035 - 0.049

Volume of solution = [420 + 7] = 427 mL = 0.427 L

To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

$pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}$ .......(2)

Given values:

$[HCOOK]=\frac{0.049}{0.427}$

$[HCOOH]=\frac{0.035}{0.427}$

$pK_a=3.75$

Putting values in equation 2, we get:

$pH=3.75-\log \frac{(0.049/0.427)}{(0.035/0.427)}\\\\pH=3.75-0.146\\\\pH=3.60$

Hence, the pH of the resulting solution will be 3.60

6 0

3 years ago

What type of reaction is shown below: Br2 + Kl 12 + KBr

bonufazy [111]

Answer:Chemical reaction

Explanation:

The main concept that must be applied to determine the coefficients (amount of each item) is that there must be equal amounts of each element on each side of the equation. We are not destroying or creating new atoms. In this case, the unbalanced reaction formula is:

B

r

2

+

K

I

=

K

B

r

+

I

2

There are a two problems we need to solve before it will be balanced:

There are two moles of Iodine atoms (

I

) on the right side of the equation, while there is only one mole on the right side.

There are two moles of bromine (

B

r

) atoms on the left side, while there is only one on the right.

Since there are two moles of bromine atoms on the left side, we need two moles on the right as well. We can do this by adding a coefficient of two to the

'

K

B

r

'

term in the equation. Our now modified equation looks like this:

B

r

2

+

K

I

=

2

K

B

r

+

I

2

There is one mole of Iodine atoms on the left, and two on the right. To fix this, we add a coefficient of two to the

'

K

I

'

term. The resulting equation is below.

B

r

2

+

2

K

I

=

2

K

B

r

+

I

2

Bonus step: We can also put ones in front of the coefficient-less species. This is like changing a phrase from "an apple" to "1 apple". It is the exact same thing, but makes it a little more clear sometimes. This would like like this:

1

B

r

2

+

2

K

I

=

2

K

B

r

+

1

I

2

Can you see that there is now an equal amount of each element on each side of the equation? That means that it is balanced.

5 0

3 years ago

Read 2 more answers

0.00001011 kilograms is a larger mass than 0.001011 grams. True or False

valentina_108 [34]

True...............................

6 0

3 years ago

Read 2 more answers

In which part of the cell is the majority of the energy released from the breakdown of glucose

Vladimir [108]

Answer:

Glycolysis

Explanation :

Glycolysis is the first step in the breakdown of glucose to extract energy for cell metabolism.Many living organisms carry out glycolysis as part of their metabolism. Glycolysis takes place in the cytoplasm of most prokaryotic and all eukaryotic cells.

7 0

3 years ago

Read 2 more answers