Answer:
58.32
Explanation:
The mass number is one mole and for Mg it is 24.306 for O it is 15.999 and for H it is 1.008. because O and H have 2, you multiply them by 2 and add all the numbers and it equals 58.32
Answer:
b) The dehydrated sample absorbed moisture after heating
Explanation:
a) Strong initial heating caused some of the hydrate sample to splatter out.
This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).
b) The dehydrated sample absorbed moisture after heating.
Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.
c) The amount of the hydrate sample used was too small.
It will create some errors but they do not create a difference of 13% difference as stated in the problem.
d) The crucible was not heated to constant mass before use.
Here the error is small.
e) Excess heating caused the dehydrated sample to decompose.
Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.
Answer:
Option B. 10
Explanation:
If 1 mol of butanol contains 6×10²⁴ atoms of H, let's calculate the amount of H.
(number of atoms / NA)
6.02 x 10²³ atoms ___ 1 mol
6×10²⁴ atoms will occupy (6×10²⁴ / NA) = 9.96 moles
H, has 10 moles in the butano formula.
True, wetlands help to prevent flooding.
Answer:
Sc
:
4s
2
3d
1
Y
:
l
5s
2
4d
1
La
:
6s
2
5d
1
Ce
:
6s
2
4f
1
5d
1
Gd
:
6s
2
4f
7
5d
1
Lu
:
6s
2
4f
14
5d
1
Ac
:
7s
2
6d
1
Pa
:
7s
2
5f
2
6d
1
U
:
l
7s
2
5f
3
6d
1
Np
:
7s
2
5f
4
6d
1
Cm
:
7s
2
5f
7
6d
1
b. s2p3
he pnictogens:
N
l
:
2s
2
2p
3
P
l
:
3s
2
3p
3
As
:
4s
2
3d
10
4p
3
Sb
:
5s
2
4d
10
5p
3
Bi
:
6s
2
4f
14
5d
10
6p
3
Mc
:
7s
2
5f
14
6d
10
7p
3
c.The noble gases:
Ne
:
2s
2
2p
6
Ar
:
3s
2
3p
6
Kr
:
4s
2
3d
10
4p
6
Xe
:
5s
2
4d
10
5p
6
Rn
:
6s
2
4f
14
5d
10
6p
6
Og
:
7s
2
5f
14
6d
10
7p
6
Explanation: From the periodic tables we can drive elements with the electronic configuration
Sc
:
4s
2
3d
1
Y
:
l
5s
2
4d
1
La
:
6s
2
5d
1
Ce
:
6s
2
4f
1
5d
1
Gd
:
6s
2
4f
7
5d
1
Lu
:
6s
2
4f
14
5d
1
Ac
:
7s
2
6d
1
Pa
:
7s
2
5f
2
6d
1
U
:
l
7s
2
5f
3
6d
1
Np
:
7s
2
5f
4
6d
1
Cm
:
7s
2
5f
7
6d
1
b. s2p3
he pnictogens:
N
l
:
2s
2
2p
3
P
l
:
3s
2
3p
3
As
:
4s
2
3d
10
4p
3
Sb
:
5s
2
4d
10
5p
3
Bi
:
6s
2
4f
14
5d
10
6p
3
Mc
:
7s
2
5f
14
6d
10
7p
3
c.The noble gases:
Ne
:
2s
2
2p
6
Ar
:
3s
2
3p
6
Kr
:
4s
2
3d
10
4p
6
Xe
:
5s
2
4d
10
5p
6
Rn
:
6s
2
4f
14
5d
10
6p
6
Og
:
7s
2
5f
14
6d
10
7p
6
