Answer:
=> 1366.120 g/mL.
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 25 Kg
Volume (v) = 18.3 mL.
From our question, we are to determine the density (rho) of the rock.
The formula:

First let's convert 25 Kg to g;
1 Kg = 1000 g
25 Kg = ?

= 25000 g
Substitute the values into the formula:

= 1366.120 g/mL.
Therefore, the density (rho) of the rock is 1366.120 g/mL.
Answer:
i cant see it very much i dont think no one can
Explanation:
Answer:
35Cl = 75.9 %
37Cl = 24.1 %
Explanation:
Step 1: Data given
The relative atomic mass of Chlorine = 35.45 amu
Mass of the isotopes:
35Cl = 34.96885269 amu
37Cl = 36.96590258 amu
Step 2: Calculate percentage abundance
35.45 = x*34.96885269 + y*36.96590258
x+y = 1 x = 1-y
35.45 = (1-y)*34.96885269 + y*36.96590258
35.45 = 34.96885269 - 34.96885269y +36.96590258y
0.48114731 = 1,99704989y
y = 0.241 = 24.1 %
35Cl = 34.96885269 amu = 75.9 %
37Cl = 36.96590258 amu = 24.1 %
When they ask you for the solution they are usually asking for x. So solve for X.
X=-12 so you only have one solution. so the answer is B
Answer:
semipermeability
Explanation:
partially but not freely or wholly permeable specifically : permeable to some usually small molecules but not to other usually larger particles a semipermeable membrane.