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Mashcka [7]
3 years ago
5

The student plans to conduct a spectrophotometric analysis to determine the concentration of Cu2+(aq) in a solution. The solutio

n has a small amount of Co(NO3)2 (aq) present as a contaminant. The student is given a diagram below, which shows the absorbance curves for aqueous solutions of Co2+ (aq) and Cu2+ (aq)
c. The spectrophotometer available to the student has a wavelength range of 400 nm to 700 nm. What wavelength should the student use to minimize the interference from the presence of the Co2+ (aq)?

Chemistry
1 answer:
mestny [16]3 years ago
7 0

Answer:

The wavelength the student should use is 700 nm.

Explanation:

Attached below you can find the diagram I found for this question elsewhere.

Because the idea is to minimize the interference of the Co⁺²(aq) species, we should <u>choose a wavelength in which its absorbance is minimum</u>.

At 400 nm Co⁺²(aq) shows no absorbance, however neither does Cu⁺²(aq). While at 700 nm Co⁺²(aq) shows no absorbance and Cu⁺²(aq) does.

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What two types of elements are ionic compounds made of?
mash [69]

Answer:

ionic compounds are made from metal and non mental elements

Explanation:

in this case it is know as an ionic compound because it contains a charge. For example, NaCl is simply a compound as it contains no charges (the charges cancel out as Cl is -1 and Na is +1)

but OH- is an ionic compound as it has a charge if -1 (O has a -2 charge and H has a +1 charge, so -2+1=-1 so OH has -1 charge)

8 0
3 years ago
How many anions are in 0.500 g of MgBr2
Oksana_A [137]

<u>Given:</u>

Mass of MgBr2 = 0.500 g

<u>To determine:</u>

Number of anions in 0.500 g MgBr2

<u>Explanation:</u>

Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol

Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles

Based on stoichiometry-

1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions

Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-

Now,

1 mole of Br- contains 6.023 * 10²³ anions

0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions

Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2


4 0
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Answer:

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Explanation:

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Answer:

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Explanation:

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