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2 years ago

What volume of 0.205 M K3PO4 solution is necessary to completely react with 134 mL of 0.0102 M NiCl2?

1 answer:

5 0

the balanced equation for the above reaction is as follows;
3NiCl₂ + 2K₃PO₄ ---> Ni₃(PO₄)₂ + 6KCl
stoichiometry of NiCl₂ to K₃PO₄ is 3:2
the number of NiCl₂ moles reacted - 0.0102 mol/L x 0.134 L = 0.00137 mol
according to molar ratio
if 3 mol of NiCl₂ reacts with 2 mol of K₃PO₄
then 0.00137 mol of NiCl₂ reacts with - 2/3 x 0.00137 = 0.000911 mol of K₃PO₄

molarity of given K₃PO₄ solution - 0.205 M
there are 0.205 mol in 1000 mL
therefore volume of 0.000911 mol - 0.000911 mol / 0.205 mol/L = 4.44 mL
volume of K₃PO₄ needed is 4.44 mL

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Which pair of reactants will undergo aldol condensation to produce methylvinyl ketone?

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When formaldehyde and acetone then react with each other( aldol condensation) then it will be formed methyl vinyl ketone.

In organic chemistry, an aldol condensation would be a condensation reaction in which an enol and enolate ion combines with a carbonyl chemical to produce a -hydroxy aldehyde or -hydroxy ketone, that is then dehydrated to produce a conjugated enone.

In aldol condensation, when formaldehyde and acetone then react with each other then it will be formed methyl vinyl ketone.

It can be written as

$CH_{2} O + CH_{3} COCH_{3}$ → $HOCH_{2} -CH_{2} -CO-CH_{3}$

When it will be heated then it gives methyl vinyl ketones.

$HOCH_{2} -CH_{2} -CO-CH_{3}$ → $CH_{2} =CH-CO-CH_{3}$

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1 year ago

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3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

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3 years ago