Hello,
I believe this is (25 g Al) * (1 mol H / 1.008 g H) = 24.8015 mol H
Hopefully that was the right approach.
I thought that if one mole of Al is
26.982 g, then the numbers of moles for H (1.008 g/mol) would be about 25 mol.
If it doesn’t make sense or if you have resources/notes to which I can refer, I’ll see what I can do!
Good luck!
Answer:
0; +1; -1
Explanation:
The resonance structure of HN₃ is shown below (you can also use horizontal dashes to represent the bonding pairs).
The molecule has 16 valence electrons, and each N atom has an octet.
To get the formal charges, cut the covalent bonds in half.
Each atom gets the electrons on its side of the cut.
Formal charge = valence electrons in isolated atom - electrons on bonded atom
FC = VE - BE
(a) On Nₐ
VE = 5
BE = 1 lone pair (2)+ 3 bonding electrons = 2 + 3 = 5
FC = 5 - 5 = 0.
(b) On Nb:
VE = 5
BE = 4 bonding electrons = 4
FC = 5 - 4 = +1
(c) On Nc:
VE = 6
BE = 2 lone pairs(4) + 2 bonding electrons = 4 + 2 = 6
FC = 5 - 6 = -1
Answer: Convection Currents
Explanation: The mantle rock is solid and denser because it is cooler. This means it will sink closer to lower mantle as warmer, lense dense rock/magma moves upwards.
Answer:
4.31 × 10²
Explanation:
Equation of the reaction;
⇌ 
The ICE Table is shown as follows:
⇌
Initial 3.10 2.50 0
Change - x -x + 2x
Equilibrium (3.10 - x) 0.0800 2x
From ;
We can see that 2.50 - x = 0.0800
So; we can solve for x;
x = 2.50 - 0.0800
x = 2.42
which = (3.10 -x) will be :
= 3.10 - 2.42
= 0.68
= 2x
= 2 (2.42)
= 4.84
![K_c = \frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
430.62
≅ 431
= 4.31 × 10²
Let the total mass of compound is 100g
The mass of each element will be
Al = 22.10 g
P = 25.40 g
O = 52.50 g
In order to determine the molecular formula we will calculate the molar ratio of the given elements
Atomic weight of Al : 27 g/ mol
Atomic weight of P : 3 1g /mol
Atomic weight of O : 16 g /mol
Moles of Al = mass / atomic mass = 22.10 / 27 = 0.819
Moles of P = mass / atomic mass = 25.40/ 31 = 0.819
Moles of O = mass / atomic mass = 52.50/ 16 = 3.28
Now we will divide the moles of each element with the lowest moles obtained to obtain a whole number ratio of moles of each element present
moles of Al = 0.819 / 0.819 = 1
moles of P = 0.819 / 0.819 = 1
moles of O = 3.28 / 0.819 = 4
So the empirical formula will be : AlPO4
