Answer:
A. Quantitative data.
Explanation:
Quantitative data involves specific numbers, while qualitative verbally describes something.
<span>361.4 pm is the length of the edge of the unit cell.
First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight.
Atomic weight copper = 63.546
Now divide the mass by the atomic weight, getting
8.94 g / 63.546 g/mol = 0.140685488 mol
And multiply by Avogadro's number to get the number of atoms:
0.140685488 * 6.022140857x10^23 = 8.472278233x10^22
Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume.
8.472278233x10^22 / 4 = 2.118069558x10^22
Now calculate the volume each unit cell occupies.
1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3
Let's get the cube root to get the length of an edge.
(4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm
Now let's convert from cm to pm.
3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm
Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.</span>
Idk you didn't be very specific
Answer:
Used for liquid extractions
Explanation:
Explanation:
The reaction equation will be as follows.
Calculate the amount of dissolved as follows.
It is given that = 0.032 M/atm and = atm.
Hence, will be calculated as follows.
=
=
=
or, =
It is given that
As,
=
=
Since, we know that pH =
So, pH =
= 5.7
Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.