A 10.0 g gold ring with a specific heat 0.129 at 24.00°C is placed in a calorimeter with 118 g of water at 1.00°C. What will be the final temperature of the system?
• Round your answer to two decimal place.
Use 4.184 - for the specific heat of water.
gº

idk

explanation:

your parents i think this could be it

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally