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Fed [463]
2 years ago
14

A cabbage p h indicator color key. entries are 0 to 2, bright red; 3 to 4, pink; 5 to 6, purple; 7, dark violet; 8 to 9, blue; 1

0 to 11, bright green; 12 to 13, lighter green. why would you use a solution, such as a cabbage ph indicator, to measure the ph of household items? to see if food went bad to test the safety of water to make sure conditions are safe to use up old cabbage
Chemistry
1 answer:
Ira Lisetskai [31]2 years ago
6 0

We use a solution, such as a cabbage ph indicator, to measure the ph of household items :

  • to see if food went bad
  • to test the safety of water
  • to make sure conditions are safe

<h3>Cabbage ph indicator:</h3>

Anthocyanin, a water-soluble pigment found in red cabbage, can change color when combined with an acid or a basic. The color changes to red in situations with a pH under 7, and to bluish-green in alkaline (basic) environments with a pH above 7. Cabbage isn't known to cause acid reflux, but plenty of other common foods are.

Anthocyanin turns pink when it interacts with the hydronium ions in an acid, and blue or green when it interacts with the hydroxide ions in a base. Because it may inform us if a substance is acidic or basic by changing color, cabbage juice is referred to as a pH indicator.

Learn more about cabbage pH here:

brainly.com/question/9626174

#SPJ4

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The empirical formula is C2H4O3

The molecular formula is C4H8O6

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Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈  2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

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3 years ago
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