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tensa zangetsu [6.8K]
1 year ago
9

A voltaic cell that uses the reaction

Chemistry
1 answer:
love history [14]1 year ago
4 0
Did you intend to write [PdCl4]^-2 instead of PdCl2-4? If so, then: 

<span>Cathode: [PdCl4]^-2(aq) + 2e- ======⇒ Pd(s) + 4Cl-(aq) </span>

<span>Anode: Cd(s) ==⇒ Cd+2(aq) + 2e-</span>
You might be interested in
In the PhET simulation, make sure that the checkbox Stable/Unstable in the bottom right is checked. Using the PhET simulation as
lord [1]

Answer:

kindly check the EXPLANATION SECTION

Explanation:

In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio  of any nuclei ranges from  to 1 to 1.5, then we say the nuclei is STABLE.

Also, we need to understand that when the  Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.

So, let us check which is stable and which is unstable:

a. 4 protons and 5 neutrons =  Neutron- proton ratio = N/P = 5/4= stable.

b. 7 protons and 7 neutrons  =  Neutron- proton ratio = N/P = 7/7= 1 = stable.

c. 2 protons and 3 neutrons  =  Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.

d. 3 protons and 0 neutrons  =  Neutron- proton ratio = N/P = 0/3= 0= unstable.

e. 6 protons and 5 neutrons  =  Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.

f. 9 protons and 9 neutrons  =  Neutron- proton ratio = N/P = 9/9 = 1 = stable.

g. 8 protons and 7 neutrons  =  Neutron- proton ratio = N/P =  7/8 =0.875 = unstable.

h. 1 proton and 0 neutrons =  Neutron- proton ratio = N/P = 0/1 =0 = unstable

6 0
2 years ago
Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As
Crank

Answer:

1.8 × 10⁻¹⁶ mol  

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

                         1.2 + 3s          2s

K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} =  1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0
2 years ago
A man pours some hot coffee into a small cup from a large jug that contains enough coffee for several cups. Which statement corr
antiseptic1488 [7]
I would say the answer is C) because the thermal energy of the coffee going into the cup has to level out until the cup gets warmer and the coffee gets cooler, and they reach the same temperature.Meaning that the jug which has already been at the same temperature(we can assume based on the verbiage “the coffee was poured from the jug to the cup”) which would give the jug more thermal energy.
8 0
1 year ago
Read 2 more answers
the image above shows a chamber with a fixed volume filled with gas at a pressure of 1560 mmHg and a temperature of 445.0 K. If
Sedaia [141]

Answer:

The new pressure of the gas in the chamber is 1,093.75 mmHg

Explanation:

The Gay-Lussac Law is a gas law that relates pressure and temperature to constant volume. This law says that the pressure of the gas is directly proportional to its temperature.

That is, if the temperature increases, the pressure increases, while if the temperature decreases, the pressure decreases. So the Gay-Lussac law can be expressed mathematically as follows:

\frac{P}{T} =k

Having an initial and an end state of a gas, the following expression can be used:

\frac{P1}{T1} =\frac{P2}{T2}

In this case:

  • P1= 1560 mmHg
  • T1= 445 K
  • P2=?
  • T2= 312 K

Replacing:

\frac{1560 mmHg}{445 K} =\frac{P2}{312 K}

Solving:

P2=\frac{1560 mmHg}{445 K} *312K

P2=1,093.75 mmHg

<u><em>The new pressure of the gas in the chamber is 1,093.75 mmHg</em></u>

7 0
1 year ago
Oxyacetylene torches produce such high temperature that they are often used to weld and cut metal. When 1.53 g of acetylene (C2H
hodyreva [135]

Answer: 4403kJ/mole

Explanation:

First we have to calculate the heat absorbed by bomb calorimeter  

Formula used :

q_b=c_b\times (T_{final}-T_{initial})

q_b = heat absorbed by calorimeter = ?

c_b = specific heat of = 10.69 kJ/K

T_{final} = final temperature = 44.688^oC=(273+44.688)K=317.688K

T_{initial} = initial temperature = 20.486^oC=(273+20.486)K=293.486K

q_b=10.69kJ/K\times (317.688-293.486)=258.7kJ

As heat absorbed by calorimeter is equal to the heat released by acetylene during combustion.

Thus 1.53 gram of acetylene releases heat of combustion = 258.7kJ

So, 26.04 g/mole of acetylene releases heat of combustion \frac{258.7}{1.53}\times 26.04=4403kJ/mole

Therefore, the heat of combustion of acetylene is, 4403kJ/mole

7 0
2 years ago
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