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Romashka [77]
2 years ago
5

You find a mysterious white powder in your kitchen. It could be cream of tartar (pH=5), sugar (pH=7), baking soda (pH=8), or dra

in cleaner (pH=14). Explain which pH indicator(s) you would use to determine the unknown substance.
1) Phenolphthalein (PHT) 2) Red cabbage 3) Bromthymol blue 4)Congo red (CR)
Chemistry
1 answer:
GREYUIT [131]2 years ago
8 0

The pH indicators to be used are Phenolphthalein, Red cabbage, Bromthymol blue and Congo red.

<h3>What are pH indicators?</h3>

Indicators are substances which change color as the pH of a medium changes.

The common indicators and their pH range is as follows:

  • Phenolphthalein - pH range of 8.3 and 10.5
  • Red cabbage - pH 2 to 10
  • Bromthymol blue - 6.0 to 7.6
  • Congo red - 3.0 to 5.2

Therefore, the indicators to be used are Phenolphthalein, Red cabbage, Bromthymol blue and Congo red.

Learn more about pH indicators at: brainly.com/question/13779537

#SPJ1

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Answer:

Explanation:

The following are the order of point from oldest to most recent

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4 years ago
The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Usi
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Answer:

\Delta H^0 _{reaction} = - 54.04 \ kJ/mol

Explanation:

The given equation for the chemical reaction can be expressed as;

2H_2O_{(l)}  \to 2H_2O_{(g)} +  O_{2(g)}

Using Hess Law to determine how much heat is produced by the decomposition of exactly 1 mole of H2O2 under standard conditions; we have the expression showing the Hess Law as follows:

\Delta H^0 _{reaction} = \sum n* \Delta H^0 _{products} -  \sum n* \Delta H^0 _{reactants}

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the molar enthalpies of the given equation are as follows:

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\Delta H_  O_{2(g)} = 0 \ kJ/mol

\Delta H _{H_2O_{(l)}}= -187.78  \ kJ/mol

Replacing them into above formula; we have:

\Delta H^0 _{reaction} = (2*(-241.82\ kJ/mol) + 0 \ kJ/mol + (2 *(-187.78 \ kJ/mol))

\Delta H^0 _{reaction} =-108.08 \ kJ/mol

The above is the amount of heat of formation for two moles of hydrogen peroxide; thus for 1 mole hydrogen peroxide ; we have :

\Delta H^0 _{reaction} = \dfrac{-108.08 \ kJ/mol}{2}

\Delta H^0 _{reaction} = - 54.04 \ kJ/mol

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