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WARRIOR [948]
3 years ago
6

Chemistry help. I need help with an assignment due in 20 minutes.

Chemistry
1 answer:
jasenka [17]3 years ago
4 0

Answer:

wheres the pic of it?

Explanation:

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D. CI706

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EXPERIMENT: THE CABBAGE INDICATOR
Nikitich [7]

Answer:

EXPERIMENT: THE CABBAGE INDICATOR

Here are your goals for this lesson:

Use indicator for each substance to determine if substance is an acid or a base

Summarize results and the experimental method

You can use the natural indicator in red cabbage juice to determine which household liquids are acids or bases. If no change in the color of the cabbage juice takes place, the liquid is neither an acid nor base; it is neutral.

Acids and bases are opposites. When an acids and base are mixed, they neutralize each other. You can use an indicator to see the neutralized reaction. The cabbage juice will change color when the liquids are neutralized.

Online Lab

This video will demonstrate how an indicator, in this case, cabbage indicator, can be used to determine whether a solution is acidic or basic. As you watch the video, remember to record your data and observations to use to present your findings.

Compile a summary of your findings from this experiment. Include your hypothesis, observations, data, and conclusions. Be sure to answer the questions below as well as explaining the method and results.

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Explanation:

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A student determines the aluminum content of a solution by first precipitating it as aluminum hydroxide, and then decomposing th
Vladimir [108]

Answer: The student should obtain <u>1.103 g of aluminum oxide </u>

Explanation:

  • First we write down the equations that represent the aluminum hydroxide precipitation  from the reaction between the aluminum nitrate and the sodium hydroxide:

Al(NO3)3 + 3NaOH → 3NaNO3 + Al(OH)3

Now,  the equation that represents the decomposition of the hydroxide to aluminum oxide by heating it.

2Al(OH)3 → Al2O3 + 3H2O

  • Second, we gather the information what we are going to use in our calculations.

Volumen of  Al(NO3)3 = 40mL

Molar concentration of Al(NO3)3 =  0.541M

Molecular Weight Al2O3 = 101.96 g/mol

  • Third, we start using the molar concentration of the aluminum nitrate and volume used to find out the total amount of moles that are reacting

\frac{0.541moles Al(NO3)3}{1L} x\frac{1L}{1000mL} x 40mL Al(NO3)3 =  0.022moles Al(NO3)3

then we use the molar coefficients from the  equations to discover the amount of Al2O3  moles produced

0.022moles Al(NO3)3 x \frac{1mol Al(OH)3}{1mol Al(NO3)3} X\frac{1mol Al2O3}{2molAl(OH)3} = 0.011 moles Al2O3

finally, we use the molecular weight of the Al2O3  to calculate the final mass produced.

0.011moles Al2O3 x \frac{101.96g Al2O3}{1mol Al2O3} = 1.103g Al2O3

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