SOMEONE ANSWER THIS FAST ITS DUE TODAY A miner finds a small minerallsragment with a volume of 11.5 cm3 and a mass of 57.4 g.
What is the density of the mineral fragment?
0.20 g/cm3
5.00 g/cm
5.74 g/cm
164 g/cm3

Density = mass/ volume

Explanation:

57.4g / 11.5cm3 = 4.991304(g/cm)

round to sig figs = 5.00(g/cm)

pie chart because it is in percents ( % )

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