The correct answer would be that substance A is rice, and substance B is sugar.
For substance A to be filtered out, it means that A is a water-insoluble substance and for substance B to be recoverable by evaporating off the water, it means that B is water-soluble.
Thus, A can neither be sugar nor alcohol because neither can be recovered by filtration. Also, B can neither be coffee nor alcohol because they can not be recovered as residue through evaporation of water.
Therefore, the only correct answer is that substance A is rice, and substance B is sugar because rice can be filtered out from water while the water-soluble sugar can be recovered by evaporating the water.
More about separating mixtures here: brainly.com/question/2827008
Answer:
Homologous pairs of cells are present in meiosis I and separate into chromosomes before meiosis II. In meiosis II, these chromosomes are further separated into sister chromatids. Meiosis I includes crossing over or recombination of genetic material between chromosome pairs, while meiosis II does not.
Explanation:
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I think the correct answer from the choices listed above is the second option. The enzyme pepsin that is present in the stomach denature in the intestine due to the alkaline environment of the intestine. This is because pepsin has an optimum pH that is very low which is ideal for the acidic conditions of the stomach.
