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Igoryamba
2 years ago
15

Noclouttrio -inta ………………….!.!.!!..!

Chemistry
1 answer:
Ronch [10]2 years ago
4 0

Answer:

3 more brlst

Explanation:

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Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
inn [45]

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

4 0
3 years ago
Riddle: What has to be broken before you can use it?
pickupchik [31]

Answer:

An egg will be your ans because it needs to be broken before we can cook it or use it for making other delicases

Explanation:

<em><u>Hope </u></em><em><u>it </u></em><em><u>helps </u></em>

4 0
1 year ago
(Science)<br> why is the 10g of shampoo is more dense than the 10kg of the same shampoo?
igor_vitrenko [27]
Grams (g) is much lighter than kilograms (kg)
6 0
2 years ago
A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ
NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂   (1)        

b) The precipitate formed is Ni(OH)₂  

 

c) The limiting reactant is:

n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles

n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

n = \frac{2}{1}*0.030 moles = 0.060 moles                  

Hence, the limiting reactant is KOH.  

d) The mass of the precipitate formed is:

n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles

m = n*M = 0.010 moles*92.72 g/mol = 0.927 g  

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M  

C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M

C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M

I hope it helps you!                                                                        

5 0
3 years ago
Which has the greater polarizing power, Li+ or Be2+ and why?
kvasek [131]

Answer:

Polarizing power refers to an atoms ability to pull an electron toward it, polarizing the atom the electron comes from. Since cations are positive, they are able to attract electrons toward themselves. Anions are negative and so do not attract more electrons.

Therefore, Be2+ has a higher polarizing power because it has a higher quantitiy of protons, hence a higher polarizing power.

4 0
3 years ago
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