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N76 [4]
2 years ago
13

Why were the Spanish interested in conquering Peru?

Chemistry
1 answer:
mr_godi [17]2 years ago
6 0

Answer:

531–1532 – Pizarro's third voyage to Peru. Spaniards form a bond with the Natives (Huancas, Chankas, Cañaris and Chachapoyas) who were under the oppression of the Inca Empire, and Pizarro includes them among his troops to face the Incas. Atahualpa is captured by Spanish.

Explanation:  Is this right ?

if it is please give thanks

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A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of
Ratling [72]

Answer:

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

Explanation:

<u>Step 1: </u>Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2-      Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

<u>For copper</u>  we have:  Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

3 0
3 years ago
What is the charge on Pb in Pb(SO3)2
Maslowich
The charge on Pb in Pb(SO3)2 is Lead (IV) Sulfite.
7 0
3 years ago
Find the potentials of the following electrochemical cell:
melomori [17]

Answer: 0.18 V

Explanation:-

Cd/Cd^{2+}(0.10M)//Ni^{2+}(0.50M)?Ni

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

E^0_(Cd^{2+}/Cd)=-0.40V[/tex]

E^0_(Ni^{2+}/Ni)=-0.24V[/tex]

Cd+Ni^{2+}\rightarrow Cd^{2+}+Ni

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}

E^0=-0.24-(-0.40)=0.16V

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential = 0.16 V

E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}

E_{cell}=0.18V

Thus the potential of the following electrochemical cell is 0.18 V.

6 0
3 years ago
Choose all the answers that apply.
mina [271]
Is created by the movement of electrons
6 0
3 years ago
How many grams of copper
iren [92.7K]
  • <u>The</u><u> </u><u>mas</u><u> </u><u>of copper</u><u> </u><u>is</u><u> </u><u>217</u><u>.</u><u>3</u><u> </u><u>grams</u><u>.</u>
7 0
2 years ago
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