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2 years ago

Draw the diazonium cation formed when cytosine reacts with nano2 in the presence of hcl.

1 answer:

7 0

Answer in the Word document below.
Diazonium compounds are a group of organic compounds sharing a common functional group R−N₂⁺. The process of forming diazonium compounds is called diazotation and usually are prepared by treatment of aromatic amines with nitrous acid and additional acid (hydrochloric acid).
Cytosine is one of the four main bases found in DNA and RNA.

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A line in the Brackett series of hydrogen has a wavelength of 1945 nm. From what state did the electron originate?

fiasKO [112]

Rydberg Eqn is given as:
1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 2626 nm = 2.626×10^-6 m 
R is the Rydberg constant: R = 1.09737×10^7 m-1 
From Brackett series n1 = 4 
Hence 1/(2.626×10^-6 ) = 1.09737× 10^7 [1/4^2 – 1/n2^2] 
Some rearranging and collecting up terms: 
1 = (2.626×10^-6)×(1.09737× 10^7)[1/16 -1/n2^2] 
1= 28.82[1/16 – 1/n2^2] 
28.82/n^2 = 1.8011 – 1 = 0.8011 
n^2 = 28.82/0.8011 = 35.98 
n = √(35.98) = 6

4 0

2 years ago

Read 2 more answers

The human body is 0.0040% iron. How many milligrams of iron does a 165 pound person contain?

Fofino [41]

Answer:

in a body of 165 pounds, the iron content is 2993.707 mg

Explanation:

% iron = 0.0040% = (mass iron / mass human) * 100

⇒ mass iron / mass human = 4 E-5

∴ mass human = 165 pounds

⇒ mass iron = 165 pounds * 4 E-5

⇒ mass iron = 6.6 E-3 pounds * ( 453.592 g/pound ) * ( 1000 mg/g )

⇒ mass iron = 2993.707 mg iron

8 0

2 years ago

How many grams of CaF2 are present in 1.25 L of a 0.15 M solution of CaF2? How do I find the grams I am confused on that part?

dybincka [34]

Answer:

Mass = 14.64 g

Explanation:

Given data:

Volume of solution = 1.25 L

Molarity of Solution = 0.15 M

Mass of CaF₂ = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

We will calculate the number of moles of CaF₂ and then determine the mass by using number of moles.

0.15 M = number of moles of solute / 1.25 L

number of moles of solute = 0.15 M × 1.25 L

number of moles of solute = 0.1875 mol/L × L

number of moles of solute = 0.1875 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.1875 mol ×78.07 g/mol

Mass = 14.64 g

7 0

2 years ago

Insoluble sulfide compounds are generally black in color. Which of the following combinations could yield a black precipitate? C

Lubov Fominskaja [6]

Answer:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

Explanation:

For these reactions, the products are:

1. Na₂S(aq) + 2 KCl(aq) → K₂S + 2 NaCl

2. Li₂S(aq) + Pb(NO₃)₂(aq) → PbS + 2 Li(NO₃)

3. Pb(ClO₃)₂(aq) + 2 NaNO₃(aq) → Pb(NO₃)₂ + 2 Na(ClO₃)

4. AgNO₃(aq) + KCl(aq) → AgCl + KNO₃

5. 2 K₂S(aq) + Sn(NO₃)₄(aq) → SnS₂ + 4 KNO₃

K₂S is a soluble sulfide that, in solid state, is white. As K has a low electronegativity, the relative polar bond K-S allows its dissolution in water.

PbS is a black and insoluble compound. As Pb electronegativity is higher than K electronegativity, The bond is less-polar and its dissolution will not be allowed.

For the third reaction, all nitrates are soluble and Na(ClO₃) is also a very soluble compound.

AgCl is an insoluble white compound.

As Sn electronegativity is high, the solubility of this compound is very low. Also, this compound is black!

Thus, combinations could yield a black precipitate are:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

I hope it helps!

6 0

2 years ago

Zinc sulfide reacts with oxygen according to the reaction 2ZnS(s)+3O2(g)→2ZnO(s)+2SO2(g) A reaction mixture initially contains 5

LenKa [72]

Answer:

The answer to your question is: Excess oxygen = 2.3 mol

Explanation:

Data

ZnS = 5 mol

O2 = 9.8 mol

Excess reactant = ?

Balanced reaction

2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)

MW ZnS = 65 + 32 = 97 x 2 = 194 g

MW O2 = 16 x 6 = 96 g

2 mol of ZnS ------------------- 3 mol O2

Ratio from the reaction = 3 mol O2/ 2 mol ZnS

= 1.5

Ratio from the quantities in the experiment = 9.8 mol O2 / 5 mol of ZnS

= 1.96

Excess reactant = Oxygen because the ratio increases

2 mol of ZnS ------------------- 3 mol O2

5 mol of ZnS ------------------- x

x = (5 x 3) / 2

x = 7.5 mol of O2

Excess Oxygen = 9.8 mol - 7.5 mol

Excess oxygen = 2.3 mol

4 0

2 years ago