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Digiron [165]
4 years ago
8

What happens to the area of the world off the coast of Peru

Chemistry
1 answer:
joja [24]4 years ago
5 0

Answer:The Peruvian west coast bordering the Pacific Ocean is a long desert strip that stretches from the departments of Tumbes in the north bordering Ecuador, to Tacna in the south bordering Chile for a total length of 1,555 miles or 2,500 km. The average temperature from December to April is 25 to 28C or 53 to 59F and the average temperature from May to November is 12 to 15C or 53 to 59F.

The coast covers about 10% of the territory but is home to more than 50% of the population. Large cities such as Lima, the capital, Trujillo, Chiclayo and Tacna are located in the coast. Cities along the coast are the center of economic activity attracting people from the interior of the country looking for economic prosperity.

idk hope this helps !!!!

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What is oxidized in a galvanic cell with aluminum and gold electrodes ?
k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V

Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Notice that the overall cell potential upon summing is:

E_{cell}=1.66 V + 1.50 V=3.16 V

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0
3 years ago
If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0
3 years ago
How many hours are in 25 years?
Annette [7]
There are *219,000* hours in 25 years
6 0
3 years ago
Read 2 more answers
Which is the MOST precise measurement this cylinder gives for the volume of liquid it contains?
disa [49]

Hello! Before I answer to your question, please be sure to include a picture or else a report will be filed for your question would be commenced. You are lucky I have this for my homework tonight and I figured it out. Thank you:

The answer to your question would be as followed:

The most precise measurement for the cylinder you are mentioning is <em><u>B. 43.0mL</u></em>

6 0
3 years ago
Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N (Kb = 5.2 × 10−4
Nataliya [291]

Answer:

(a) 10.62

(b) 2.82

(c) 1.95

Explanation:

The neutralization reaction in this question is

(CH3CH2)3N + HCl   ⇒ (CH3CH2)3NH⁺ + Cl⁻

The problem  can be  solved by calculating the number of moles of triethylamine after  addition of the portions of HCl. Since it is a weak base if it is not consumed completely, that is in excess we will have a buffer of a waek base. If its consumed completely the pH will be determined by the strong acid HCl.

The pOH for a buffer of a weak base is gven by

pOH = pKb + log [(CH3CH2)3NH⁺] / [(CH3CH2)3N]

(a) 11 mL of 0.100 M HCl

mol HCl = 0.011 L x 0.100 mol/L = 0.0011 mol HCl

mol  (CH3CH2)3N reacted = 0.0011 mol

mol (CH3CH2)3NH⁺ produced = 0.0011 mol

mol (CH3CH2)3N  initially = 0.020 L x 0.1000 mol/L 0.0020 mol

mol (CH3CH2)3N left = 0.0020 mol - 0.0011 = 0.0009 mol

pKb = - log Kb = - log (5.2 x 10⁻⁴) = 3.284

Now we can compute pOH,

pOH = 3.284 + log ( 0.0011 / 0.0009 ) = 3.37

pH = 14 - pOH = 14 - 3.37 = 10.62

(b) 20.60 mL HCl

mol HCl = 0.0206 L x 0.100 mol/L = 0.00206

mol  (CH3CH2)3N consumed = 0.0020 mol

This is so  because the acid will consume completely the 0.0020 mol of the weak base  we had originally present.

Now the problem circumscribes to that of calculating the pH of the unreacted HCl

Total Vol = 0.0206 L + 0.02 L = 0.0406 L

mol HCl = 0.0206 L x .100 = 0.00206 mol

mol HCl left = 0.00206 mol - 0.0020 mol = 0.00006 mol

[HCl] = 0.00006 mol / 0.0406 L = 0.0015 M

Since HCl is a strong acid ( 100 % ionization) :

pH = - log [H⁺] = - log ( 0.0015 ) = 2.82

(c) We will compute the pH in  the same way we did for part (b)

mol HCl = 0.025 L x 0.100 mol/L = 0.0025 mol

mol HCl left = 0.0025 mol  - 0.0020 mol = 0.0005

Total Volume = 0.020 L + 0.025 L = 0.045 L

[HCl] = 0.0005 mol / 0.045 L = 0.111

pH = - log ( 0.111) = 1.95                                            

3 0
3 years ago
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