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Aleks04 [339]
3 years ago
11

Using a value of Ksp = 1.8 x 10-2 for the reaction PbCl2(s) Pb+2(aq) + 2Cl-(aq).

Chemistry
1 answer:
drek231 [11]3 years ago
7 0

Answer:

Option A and Option B

Explanation:

The concentrations of Pb+2(aq) and Cl-(aq) are expected to decrease (because you have initially very high values of [Pb2+] and [Cl-], they should decrease so that you equilibrium is reestablished, so that Qsp = Ksp).

Solids aren't measured using concentrations but the AMOUNT of PbCl2(s) is expected to increase.

It IS possible to have a system with Ksp > 1.8 x 10^-2 -- if you increase the temperature, then solubility and Ksp are both increased. It is also possible to have a system with Qsp > Ksp (as is true in this case). However, if Qsp > Ksp, the system will shift so that reequilibrium is reestablished. Over time, Qsp will be come equal to Ksp.

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A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0
bija089 [108]

Answer:

E = 0.062 V

Explanation:

(a) See the attached file for the answer

(b)

Calculating the voltage (E) using the formula;

E = - (2.303RT/nf)log Cathode/Anode

Where,

R = 8.314 J/K/mol

T = 35°C = 308 K

F- Faraday's constant = 96500 C/mol,

n = number of moles of electron = 2

Substituting, we have

E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)

   = -0.031 * -2

  = 0.062V

Therefore, the voltmeter will show a voltage of 0.062 V

5 0
3 years ago
Can someone explain the steps for balancing chemical equations in depth?
m_a_m_a [10]

Answer:

Steps explained below

Explanation:

To explain balancing of chemical equations, I will make use of an example equation where Hydrogen and oxygen react to form water.

H2 + O2 = H2O

Now, the equation I've listed above is an unbalanced chemical equation. It can be balanced by the following steps;

Step 1: Identify the elements on both the left Hand side and the right hand side.

In this case;

on the left hand side, we have H and O.

On the right hand side, we have H and I also.

Step 2: Identify the number of atoms of each element on both the left and right hand sides.

On the left, H has 2 atoms and O has 2 atoms.

On the right, H has 2 atoms and O has 1 atom.

Step 3: For the equation to be balanced, the number of atoms of each element on the right and left hand side must be the same.

Thus,

O on the left hand side has 2 atoms but on the right hand side it has 1 atom. Thus, we will multiply O on the right by 2 to balance what we have on the left.

So, we now have;

H2 + O2 = 2H2O

Step 4: Check equation: We now have;

H2 + O2 = 2H2O

Our left hand side remains 2 atoms of H and 2 atoms of O. But on the right, we now have;

2 atoms O and 4 atoms of H.

Which means atoms of H is not balanced with the left side.

Step 5: rebalance equation: To rebalance, we multiply H on the left by 2 to give us 2 × 2 = 4 atoms.

Thus, we now have;

2H2 + O2 = 2H2O

3 0
3 years ago
Read 2 more answers
When 10g of calcium carbonate is heated,4.4g of carbon dioxide escape out. The amount of residue left is:
Maslowich
The dissociation of calcium carbonate, CaCO3, to simpler compounds can be expressed as,
                        CaCO3 --> CaO + CO2
The precipitate is CaO and its amount is calculated through the difference which will give us the answer of 5.6 g. 
8 0
3 years ago
A sample of sodium-24 with an activity of 14 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24
SOVA2 [1]

Answer:

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

a)

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 =  log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

b)

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

c)

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77  mCi

d)

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

5 0
3 years ago
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
Eva8 [605]

Answer:

m_{Ag}=135.8gAg

Y=88.4\%

Explanation:

Hello there!

In this case, according to the described chemical reaction, it is possible to compute the theoretical mass of silver as mass via the 1:2 mole ratio of copper to silver and their atomic mass in the periodic table, in order to perform the following stoichiometric setup:

m_{Ag}=40.gCu*\frac{1molCu}{63.55gCu}*\frac{2molAg}{1molCu}*\frac{107.87gAg}{1molAg}\\\\   m_{Ag}=135.8gAg

Next, given the actual yield of 120 g, we compute the percent yield via:

Y=\frac{120g}{135.8g}*100\%\\\\Y=88.4\%

Regards!

4 0
3 years ago
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