from 150 gram of Urea 3 into 10 power 23 molecules are removed then how many molecules are remain

Amount remaining = 1.205 * 10²⁴ molecules of urea

Explanation:

The molecular formula of urea is CH₄N₂O. It has a molar mass of 60 g/mol.

Number of moles of urea present in 150 g of urea = mass/molar mass

Number of moles = 150 g/ 60 g/mol = 2.5 moles of urea

I mole of urea contains 6.02 * 10²³ molecules of urea

2.5 moles of urea will contain 2.5 * 6.02 * 10²³ molecules of urea = 1.505 * 10²⁴ molecules of urea

If 3 * 10²³ molecules are removed from 1.505 * 10²⁴ molecules of urea, amount remaining will be:

1.05 * 10²⁴ - 3 * 10²³ = (15.05 - 3) * 10²³ molecules of urea

Amount remaining = 12.05 * 10²³

Amount remaining = 1.205 * 10²⁴ molecules of urea

depends on if you on plato or not

cellular respiration oxidizes food into energy in the form of atps.

glycolysis (6 c hexose) > 2x atps are used > 6c hexose biphosphate > 2 x 3c triphosphate >

2 x 3c pyruvate (2x nadh produced from nad) and (4x atp produced)

pyruvate actively pumped mitochondrial matrix.