Answer:
Mass= 2.77g
Explanation:
Applying
P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28
PV=nRT
NB
Moles(n) = m/M
PV=m/M×RT
m= PVM/RT
Substitute and Simplify
m= (2.09×1.13×28)/(0.082×291)
m= 2.77g
Answer:
Conociendo el volumen de solución, masa de soluto y su masa molar, es posible determinar: B) Concentración molar
La molaridad es la relación entre el número de moles de soluto y los litros de solución. Más:
M = No moles de solución de soluto / volumen (L)
Y a su vez los moles de soluto se encuentran por:
No moles de soluto = masa soluto / masa molar soluto
Iron and steel are magnetic in nature. So you can use a magnet to separate them
Answer:
9.63 L of NO
Explanation:
We'll begin by calculating the number of mole in 50.0 g of NH₄ClO₄. This can be obtained as follow:
Mass of NH₄ClO₄ = 50 g
Molar mass of NH₄ClO₄ = 14 + (4×1) + 35.5 + (16×4)
= 14 + 4 + 35.5 + 64
= 117.5 g/mol
Mole of NH₄ClO₄ =?
Mole = mass /molar mass
Mole of NH₄ClO₄ = 50/117.5
Mole of NH₄ClO₄ = 0.43 mole
Next, we shall determine the number of mole of NO produced by the reaction of 50 g (i.e 0.43 mole) of NH₄ClO₄. This can be obtained as follow:
3Al + 3NH₄ClO₄ –> Al₂O₃ + AlCl₃ + 3NO + 6H₂O
From the balanced equation above,
3 moles of NH₄ClO₄ reacted to produce 3 moles of NO.
Therefore, 0.43 mole of NH₄ClO₄ will also react to produce 0.43 mole of NO.
Finally, we shall determine the volume occupied by 0.43 mole of NO. This can be obtained as follow:
1 mole of NO = 22.4 L
Therefore,
0.43 mole of NO = 0.43 × 22.4
0.43 mole of NO = 9.63 L
Thus, 9.63 L of NO were obtained from the reaction.
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