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DochEvi [55]
3 years ago
5

What is the application of Complexometric titration? ​

Chemistry
1 answer:
PtichkaEL [24]3 years ago
7 0

Explanation:

Complexometric titration (sometimes chelatometry) is a form of volumetric analysis in which the formation of a colored complex is used to indicate the end point of a titration. Complexometric titrations are particularly useful for the determination of a mixture of different metal ions in solution.

hope it will help u ✌️

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Draw the diazonium cation formed when cytosine reacts with nano2 in the presence of hcl.
Sindrei [870]
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 <span>are prepared by treatment of aromatic amines with </span>nitrous acid<span> and additional acid (hydrochloric acid).
</span>Cytosine is one of the four main bases found in DNA and RNA.
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7 0
3 years ago
Describe what happens when a bond is created between magnesium and bromine. Be specific and explain in terms of electrons.
Allushta [10]

Answer: Magnesium and Bromine/MgBr2 = Ionic compounds

Explanation: When atoms form together they can form between Ionic Compounds and molecules; this could depend on if they're joined by Covalent bonds as well because when atoms form with Covalent bonds, it forms Molecules.

5 0
3 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
. Who was the scientist responsible for the laws of motion?
Gennadij [26K]

Newton was the scientist

4 0
2 years ago
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Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.
Oduvanchick [21]

Answer:

103.9 g

Explanation:

  • C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:

  • 54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:

  • 1.23 mol C₃H₈ * \frac{3molCO_2}{1molC_3H_8} = 3.69 mol CO₂

We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:

  • 3.69 mol CO₂ * 44 g/mol = 162.36 g

And <u>apply the given yield</u>:

  • 162.36 g * 64.0/100 = 103.9 g
7 0
2 years ago
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