The apparent topography exhibited by minerals in thin section as a consequence of refractive index.
We are tasked to solve for the volume of Gold given that the mass of Gold is equivalent to that of a copper.
Details:
Density of Copper= 8.96 g/ml
Volume of Copper=141 ml
mass of Gold = mass of Copper
Density of Gold=19.3 g/ml
In order to solve for the mass of copper, we need to use the density formula
Density= mass/volume
Since we are to solve for the mass of copper
mass of copper= Density of Copper * Volume of Copper
mass of copper= 8.96 g/ml* 141 ml= 1263.36 g
Thus,
mass of gold=mass of copper=1263.36 g
Hence,
Using still the density formula to solve for the volume of gold,
Volume of gold=mass of gold/ Density of gold
Volume of gold=1263.36 g/ 19.3 g/ml = 65.46 mL
Therefore, the volume of the gold must be 65.46 mL in order to have the same mass of a copper
City B is at a higher temperature because as you go higher in altitude the boiling point of water becomes lower.
low melting point means that a low amount of heat is required to melt said substance.
Answer:
r = 3.61x
M/s
Explanation:
The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.
r = k.![[S2O2^{-8} ]^{x} x [I^{-} ]^{y}](https://tex.z-dn.net/?f=%5BS2O2%5E%7B-8%7D%20%5D%5E%7Bx%7D%20x%20%5BI%5E%7B-%7D%20%5D%5E%7By%7D)
K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :
x = 1
Now, to find the coefficient y let's do the same for the experiments 1 and 3:
y = 1
Now, we need to calculate the constant k in whatever experiment. Using the first :
k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]
Using the data given,
r = 
r = 3.61x M/s
