Some animals convert highly toxic `NH_(3)` into least toxic trimethylamine oxide (TMAO) and retain high concentration of TMAO and urea to minimise `H_(2)O` loss from body are

Read the following statements and find out the incorrect statements. (a) Kidney plays very significant role in the removal of NH_(3) (b) Terrestial adaptation necessitated the production of more toxic nitrogenous wastes like urea and uric acid for conservation of water . (c) Some amount of urea may be retained in the kidney matrix of some of the ureotelic animals to maintain a desired osmolarity. (d) Uricotelic animals excrete nitrogenous wastes as uric acid in the form of pellet or paste with a minimum loss of water .

With regards to toxicity and the need for dilution in water, how different are ureotelic and uricotelic excretions? Give examples of animals that use these types of excretion. Ureotelism: 1. The process of excreting urea is called ureotelism. 2.Animals which are found in places where water availability is not abundant have this mode of excretion. 3. They convert Ammonia produced in the body into urea in the liver and release it to the blood. This is filtered and excreted by the kidneys, Eg: Mammals, many terrestrial amphibians and marine fishes. 4.In terms of toxicity urea is more toxic than uric acid but it is soluble in water and is thus excreted as urine.

Fill in the blanks from the choices given in brackets- When a metallic oxide is dissolved in water, the solution formed has a high concentration of _____ions [H^(+), H_(3)O^(+), OH^(-)]

Limiting reactant: Urea [(NH_(2))_(2) CO] used as ferlilzer as animal feed, and in polymer industry, is prepared by the reaction between ammonia and carbon dioxide: 2NH_(3)(g) + CO_(2) (g) rarr (NH_(2))_(2) CO(aq.) + H_(2) O(1) In one process , 637.2g of NH_(3) is allowed to react with 11.42g of CO_(2) (i) Which of the two reactants is the limiting reactant? (ii) Calculate the mass of (NH_(2))_(2) CO formed? (iii) How much of the excess reagent (in grams) is left at the end of the reaction? Strategy: (i) Since we cannot tell by inspection which of the two recantants is the limiting reacant, we have to procced by first converting their masses into number of moles. Take each reactnat in turn and ask how many moles of product (urea) would be obtained if each were completely consumed. The reactant that gives the smaller number of moles of producet is the limiting reactant. (ii) Convert the moles of product obtained to grams of product. (iii) From the moles of product, calculate to grams fo excess reactant needed int he reaction. Then subtract this qunitity from the grams of the reactant available to find the quanity of the excess reactant remaining.