Diagram A shows the Lewis structure (LS) of CH_2O. The formal charge on each atom is zero.
To get the formal charge (FC) on the atoms, cut each bond in half, as in <em>Diagram B</em>. 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
<em>On O:
</em>
VE = 6
BE = 2 lone pairs 2 + 2 bonding electrons = 4 + 2 = 6
FC = 6 – 6 = 0.
<em>On H:
</em>
VE = 1
BE = 1 bonding electron
FC = 1 – 1 = 0
<em>On C:
</em>
VE = 4
BE = 1 in each single bond + 2 in the double bond = 2 + 2 = 4
FC = 4 - 4 = 0
Answer:
6.79 g of phosphine can be produced
Explanation:
The reaction is this:
3H₂ + 2P → 2PH₃
We have the mass of the two reactants, so let's find out the limiting reactant, so we can work with the equation. Firstly, we convert the mass to moles (mass / molar mass)
6.2 g / 30.97 g/mol = 0.200 moles of P
4g / 2 g/mol = 2 moles of H₂
Ratio is 3:2.
3 moles of hydrogen react with 2 moles of P
Then, 2 moles of H₂ would react with (2 . 2)/ 3 = 1.3 moles of P.
We have only 0.2 moles of P, so clearly the phosphorous is the limiting reactant.
Ratio is 2:2. So 2 moles of P can produce 2 moles of phosphine. Therefore, 0.2 moles of P must produce the same amount of phosphine.
Let's convert the moles to mass ( mol . molar mass)
0.2 mol . 33.97 g/mol = 6.79 g
Answer:
λ = 2.38 × 10^(-7) m
Explanation:
We are given the work function for palladium as 503.7 kJ/mol.
Now let's convert this to KJ/electron.
We know from avogadro's number that;
1 mole of electron = 6.022 × 10^(23) electrons
Thus,
503.7 kJ/mol = 503.7 × 1/(6.022 × 10^(23)) = 8.364 × 10^(-22) KJ/electron = 8.364 × 10^(-19) J/electron
Formula for energy of a photon is;
E = hv
Where;
h is Planck's constant = 6.626 × 10^(-34) J.s
v is velocity
Now, v = c/λ
Where;
c is speed of light = 3 × 10^(8) m/s
λ is wavelength of light.
Thus;
E = hc/λ
Making λ the subject, we have;
λ = hc/E
λ = (6.626 × 10^(-34) × 3 × 10^(8))/(8.364 × 10^(-19))
λ = 2.38 × 10^(-7) m
<span>
Mn²⁺ + 4H2O -----> MnO4⁻ + 8H⁺ +5e⁻ /*2
<span>NaBiO3 +6H⁺ +2e⁻ -----> Bi³⁺ + Na⁺ + 3H2O /*5
</span>2Mn²⁺ + 5 NaBiO3+8H2O+30H⁺ ---> 2MnO4⁻ +5Bi³⁺ + 5Na⁺ +16H⁺ +15H2O
</span>2Mn²⁺ + 5 NaBiO3+14H⁺ ---> 2MnO4⁻ +5Bi³⁺ + 5Na⁺ +7H2O
There are 7 water molecules in this reaction.
The complete chemical reaction of aqueous Sodium sulfate and aqueous Barium nitrate is
The right side and the left side of the reaction have equal number of elements, therefore it is balanced.
