If the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high.
Spectrometry measures the interaction of light with molecules. The absorbance refers to how much light that interacts with molecules of the substance. The more the concentration of the substance the higher the absorbance of the solution.
Hence, if the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high. An unusually high absorbance tells us that the solution is too concentrated.
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When `CO_(2)` is bubbled through a cold pasty solution of barium peroxide in water, `H_(2)O_(2)` is obtained. <br> `BaO+CO_(2)+H_(2)OtoBaCO_(3)+H_(2)O_(2)` Barium carbonate being insoluble is filtered off. This is known as Merck's process.
<h3>What is meant by Perhydrol?</h3>
perhydrol (countable and uncountable, plural perhydrols) A stabilised solution of hydrogen peroxide.
<h3>What is Merck's Perhydrol?</h3>
Uses: Perhydrol is used as an antiseptic for wounds, and also acts as a germicide to kill bacteria and germs.
Being a strong oxidizing agent it has bleaching properties and acts as a ripening agent.
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D. Being cold temperatures can result in a cold nose. With prolonged exposure The body will start to lose heat faster than it can generate it, this is the result of hypothermia.
The shells further away from the nucleus are LARGER and can hold MORE electrons
Answer:
6.78 × 10⁻³ L
Explanation:
Step 1: Write the balanced equation
Mg₃N₂(s) + 3 H₂O(g) ⇒ 3 MgO(s) + 2 NH₃(g)
Step 2: Calculate the moles corresponding to 10.2 mL (0.0102 L) of H₂O(g)
At STP, 1 mole of H₂O(g) has a volume of 22.4 L.
0.0102 L × 1 mol/22.4 L = 4.55 × 10⁻⁴ mol
Step 3: Calculate the moles of NH₃(g) formed from 4.55 × 10⁻⁴ moles of H₂O(g)
The molar ratio of H₂O to NH₃ is 3:2. The moles of NH₃ produced are 2/3 × 4.55 × 10⁻⁴ mol = 3.03 × 10⁻⁴ mol.
Step 4: Calculate the volume corresponding to 3.03 × 10⁻⁴ moles of NH₃
At STP, 1 mole of NH₃(g) has a volume of 22.4 L.
3.03 × 10⁻⁴ mol × 22.4 L/mol = 6.78 × 10⁻³ L
