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MariettaO [177]
2 years ago
11

A 25.0 mL sample of an aqueous solution of calcium hydroxide, Ca(OH)2, is titrated to the equivalence point with 21.2 mL of 0.62

0 M sulfuric acid, H2SO4, acid solution. What is the molarity of the calcium hydroxide solution?
Chemistry
1 answer:
kap26 [50]2 years ago
3 0

Let's see

  • C1V1=C2V2
  • 25C1=21.2(0.620)
  • 25C1=13.144
  • C1=0.526M
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Calculate the maximum numbers of moles and grams of iodic acid (HIO₃) that can form when 635 g of iodine trichloride reacts with
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What is Chemical Reaction?

A chemical reaction occurs when a certain group of molecules converts into another form without affecting their nuclei; only the transfer or sharing of electrons and the building and breaking of bonds occur.

Main Content

Known :

\mathrm{ICl}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+\mathrm{HCl}

Mass of ICI_3 = 635g

Mass of H_2O = 118.5g

Calculations :

First, balance the given chemical equation by place 2,3 and 5 as the coefficients of ICI_3, H_2O and HCl,\\ Respectively

2 \mathrm{ICl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+5 \mathrm{HCl}

We multiply the given mass of ICl_3 by the reciprocal of its molar mass to get the number of moles. The molar mass of ICl_3 is 233.26g/mol

\text { Moles of } \mathrm{ICl}_{3}=635 \mathrm{~g} \mathrm{ICl}_{3} \times \frac{1 \mathrm{~mol} \mathrm{ICl}_{3}}{233.26 \mathrm{~g} \mathrm{ICl}_{3}}=2.7223 \mathrm{~mol}  \mathrm{ICl}_{3}

Them, we multiply the ratio between ICl_3 and HIO_3. based on the chemical equation, the molar ratio 1 mol HIO_3/2  mol ICl_3.

\text { Moles of } \mathrm{HIO}_{3} \text { formed }=2.7223 \mathrm{~mol} \mathrm{ICl}_{3} \times \frac{1 \mathrm{~mol} \mathrm{HIO}_{3}}{2 \mathrm{~mol} \mathrm{ICl}_{3}}=1.3612 \mathrm{~mol} \mathrm{HIO}_{3}

For water, we again multiply the given mass H_2O by the reciprocal of its molar mass to get the number if moles. The molar mass of H_2O is 18.02g/mol

\text { Moles of } \mathrm{H}_{2} \mathrm{O}=118.5 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \times \frac{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{18.02 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}}=6.5760 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}

Then, we multiply the molar ratio between H2O and HIOs. Based on the chemical equation, the molar ratio is 1 mol HIO_3/3 Mole H_2O

\text { Moles of } \mathrm{HIO}_{3} \text { formed }=6.5760 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} \times \frac{1 \mathrm{~mol} \mathrm{HIO}_{3}}{3 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}=2.1920 \mathrm{~mol} \mathrm{HIO}_{3}

We can see that the limiting reactant is ICL_3 since the given mass of ICl_3 forms lesser product than water does. Thus , the maximum number of moles of HIO_3 formed is 1.36 mol HIO_3. We now multiply the molar mas of HIO_3 to the calculated number of moles. The molar mass of HIO_3 is 175.91 g/mol.

Mass of HIO_3 formed (Max) =1.3612 \mathrm{~mol} \mathrm{HIO}_{3} \times \frac{175.91 \mathrm{~g} \mathrm{HIO}_{3}}{1 \mathrm{~mol} \mathrm{HIO}}=239 \mathrm{~g} \mathrm{HIO}_{3}

We multiply the number of moles of ICl_3 by the molar ratio between ICl_3 and H_2O which is 3 mol H_2O mol ICl_3 we get the number of moles of H_2O reacted. Then, we multiply the molar mass of water.

Mass of H_2O reacted =2.7223 \mathrm{~mol} \mathrm{ICl} 3 \times \frac{3 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{2 \mathrm{~mol} \mathrm{ICl}_{3}} \times \frac{18.02 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}

Mass of H_2O reacted = 73.6g H_2O

We subtract the mass of H_2O reacted from the given mass of H_2O.

Mass of H_2O = 118.5 - 73.6g = 44.9g H_2O

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

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Answer:

Whenever the two groups on each end of a carbon-carbon double bond are the same, two diastereomers are possible.

Explanation:

Double bonds are formed when four electrons are shared between two carbon atoms. The second bond is generally known as a pi-bond while the first bond is called a sigma bond.

The presence of a double bond in a compound leads to restricted rotation about the carbon-carbon double bond. Hence the substituents are 'locked' in a particular configuration.

The more substituted a double bond is, the more stable it is according to Saytzeff's Rule.

Due to steric effects, trans alkenes are generally more thermodynamically stable than cis alkenes since trans alkenes have bulky groups oriented apart from each other.

All these are true statements. The answer selected must be the untrue statement.

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The most soluble of the compounds given above is [NH4]3PO4. [NH4]3PO4 is a salt of ammonium and phosphate; it is a white crystal at room temprature. It is an highly unstable compound and thus easily decompose when mixed with water.
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