There’s nothing to answer
Answer:
Explanation:
First you should calculate the volume of a big sphere,so:
Then you calculate the volume of a small spehre, so:
Finally you subtract the two quantities:
Answer:
See explanation
Explanation:
The oxides or hydrides are formed by exchange of valency between the two atoms involved. The group of the atom bonded to oxygen or hydrogen in the binary compound can be deduced by considering the subscript attached to the oxygen or hydrogen atom.
Now let us take the journey;
R2O3- refers to an oxide of a group 13 element, eg Al2O3
R2O - refers to an oxide of group a group 1 element e.gNa2O
RO2 - refers to an oxide of a group 14, 15 or 16 element such as CO2, NO2 or SO2
RH2 - refers to the hydride of a group 12 element Eg CaH2
R2O7 - refers to an oxide of a group 17 element E.g Cl2O7
RH3- refers to a hydride of a group 13 element E.g AlH3
Answer:
Object B has a density of 2.5 g/cm³ which is greater than object A by 1 g/cm³
Explanation:
Since we know that the formula for density is d=m/v, we can divide each mass by its corresponding volume to find the densities
12/8=1.5
20/8=2.5
So we know that object B has a greater density than object A by 1 g/cm³ (gram per cubic centimeter). Also the standard unit for density is kilograms per cubic meter but I used gram per cubic centimeter since they were the given units. 1cm=100m, 1000g=1km
Answer:
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
Explanation:
Step 1: Data given
The solution contains 0.021 M Cl- and 0.017 M I-.
Ksp(CuCl) = 1.0 × 10-6
Ksp(CuI) = 5.1 × 10-12.
Step 2: Calculate [Cu+]
Ksp(CuCl) = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [0.021]
[Cu+] = 1.0 * 10^-6 / 0.021
[Cu+] = 4.76 *10^-5 M
Ksp(CuI) = [Cu] [I]
5.1 * 10^-12 = [Cu+] [I-]
5.1 * 10^-12 =[Cu+] [0.017]
[Cu+] = 5.1 * 10^-12 / 0.017
[Cu+] = 3.0 *10^-10 M
[Cu+]from CuI hast the lowest concentration
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
