Solution: <span>1 L of solution and 1.03 g/mL = 1030 g/L </span> <span> 1030 g/L time 10.5% = 108.15 g of glucose/L</span> And <span>=108.15 g / 180.16 g/mol = 0.6003 moles in 1L </span> <span>=or 0.600 mol/kg</span>
use the equation q=mcΔt. convert 13 mg to g. so it becomes .013g which will be the mass. C is specific heat of water which is always 4.184. Delta t is 75-1 which is 74. (.013)(4.184)(74) = 4.025