You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
According to this formula:
㏑(K2/K1) = Ea/R(1/T1 - 1/T2)
when K is the rate constant
Ea is the activation energy
R is the universal gas constant
and T is the temperature K
when K is doubled so K2: K1 = 2:1 & R = 8.314 J.K^-1.mol^-1
and T1 = 10 +273 = 283 k & T2 = 21 + 273 = 294 k
So by substitution:
㏑2 =( Ea / 8.314) (1/283 - 1/294 )
∴ Ea = 43588.9 J/mol = 43.6 KJ/mol
Answer:
when energy changes from one form to another like in a hydroelectric dam that transform the kinetic energy.
Explanation:
Shape and size has no effect on an objects density