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2 years ago

Which one of the following would form a precipitate when mixed with LiOH?

Chemistry

1 answer:

ikadub [295]2 years ago

3 0

Answer:

ZnBr2

Explanation:

KNO3 + LiOH -------> no reaction

This is because KNO3 and LiOH completely ionize.in water and form neutral solutions.Since both are neutral no reaction occurs

NH4Cl + LiOH -----> NH3 (aq) + H2O (l) + LiCl(aq)

None of the above products are precipitates

Ca(C2H3O2)2 + 2LiOH -----> Ca(OH)2 (aq) + 2LiC2H3O2 (aq)

ZnBr2 + 2LiOH -----> Zn(OH)2 (s) + LiBr2 ( aq)

Zn(OH)2 thus formed is a white precipitate

However when excess LiOH is added Zn(OH)2 precipitate will dissolve to give a clear solution of Li2ZnO2 .

You can remember this by the fact that Na,K,Rb,Cs,Ca,Sr,Ba hydroxides are soluble in water and all other hydroxide are precipitated in water

Zn(OH)2 (s) + 2 LiOH ------> Li2ZnO2(aq) + 2 H2O ( l)

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Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

2 years ago

What is the volume of 500g of CO2?

Airida [17]

Weighs 0.001836 gram per cubic centimeter or 1.836 kilogram per cubic meter

Try to see if this helps

3 0

2 years ago

Read 2 more answers

Since there are many organic compounds,how can you distinguish one organic compound from another

iogann1982 [59]

What are organic and inorganic compounds? Organic chemistry is the study of the carbon compounding molecules. Inorganic chemistry, by contrast, is the study of all compounds that do NOT contain carbon compounds.

8 0

2 years ago

When chromium loses two electrons, its configuration changes to

damaskus [11]

Electronic configuration of cromium is

Cr-[Ar]4s¹3d⁵

When cromium loses two electrons it becomes Cr⁺².

So its electronic configuration becomes,

Cr⁺²-[Ar]3d⁴

One electron will go from 4s orbital and one electron will go from 3d orbital.

So the answer here is D. [Ar]3d⁴ -because after losing 2 electrons electronic configuration of cromium becomes [Ar] 3d⁴.

3 0

2 years ago

The following combinations are not allowed. If n and ml are correct, change the l value to create an allowable combination:

ololo11 [35]

The allowable combination for the atomic orbital is n=3, l=1 or 2, $m_{l}$ =+1.

<h3>What are the three quantum numbers of an atomic orbital?</h3>

Three quantum numbers specify an atomic orbital:

- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.

- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),

- $\boldsymbol{m}_{l}$ is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus; $m_{l}$ values are integers from -l to =l(l limits ml)

For n = 3, l can have three values: 0, 1, and 2. Hence, the l value must be lower than 3. Because of $m_{l}$ = +1, the l value must be higher than 0. Two l values are consistent with n and $m_{l)$ values:

l= 1 or 2

Therefore, the allowable condition is n=7, l=1 or 2, $m_{l}$ =+3.

To know more about quantum numbers, visit: brainly.com/question/16979660

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7 0

1 year ago