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3 years ago

Predict the solubility of the following substances in water and place them in the appropriate bins

1 answer:

5 0

<span> NaNO3: Soluble 2. AgBr: Insoluble 3. NH4OH: Soluble 4. Ag2CO3: Insoluble 5. NH4Br: Soluble 6. BaSO4: Insoluble 7. Pb(OH)2: Insoluble 8. PbCO3: Insoluble</span>

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I’m microsoft word you can access what from the mini toolbar

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Word count and just reall simple stuff u can just get the real version on the computer

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2 years ago

If 35.8 grams of LiCl are dissolved in 184 grams of water, what is the concentration of the solution in percent by mass?

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35.8/184=.194565, or 19.4565%

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3 years ago

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In the following reaction, which element or compound is the oxidizing agent?

8_murik_8 [283]

Answer:

$O_{2}$ is the oxidizing agent

Explanation:

An oxidizing agent is an element in a reaction that accepts the electrons of another element. It is typically hydrogen, oxide, or any halogen. In this case, it is oxygen. The answer is 02.

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3 years ago

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Based on the shapes of the water drops in part D, how does the attractive force between water and wax compare to the attractive

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Because the cohesive forces inside the droplets are stronger than the adhesive forces between both the drops and the wax, water does not penetrate waxed surfaces. Because the adhesive forces between the liquid and the glass are stronger than the cohesive forces inside the water, water wets glass and spreads out across it.

Explanation:

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1 year ago

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

3 years ago