Answer:
B
Explanation:
First of all it is important to know that a half filled orbital is particularly stable. In phosphorus all the electrons occur singly in the 3p sublevel minimizing inter electronic repulsion hence it is more difficult to remove an electron from this energetically stable arrangement. In sulphur, electrons are paired in one of the 3p orbitals thereby lowering the energy of that level due to instability caused by interelectronic repulsion between two electrons in the same orbital.
For a (unbalanced) reaction: NaOH +CO2-Na2CO3 + H2O, the moles of NaOH and moles of each product are formed are mathematically given as
a) Moles of NaOH =44.05
b) Moles of Na2CO3=21.0
<h3>What is the moles of NaOH and what moles of each product are formed?</h3>
Generally, the equation for the Chemical reaction is mathematically given as
2 NaOH(aq)+ CO2(g)------> Na2CO3(aq)+ H2O(l)
Therefore
Moles of CO2= 925/44
Moles of CO2=21.0
Hence
Moles of NaOH = 2 x Moles of CO2
Moles of NaOH = 2x925/44
Moles of NaOH =44.05
In conclusion
Moles of Na2CO3 925/44
Moles of Na2CO3=21.0
And
Moles of H2O= 925/44
Moles of H2O= 21.0
Read more about Chemical reaction
brainly.com/question/16416932
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
1 mole C3H8 produces 4 moles H2O. So, first we convert 32 grams of propane to moles and then find moles of H2O. Then convert moles of H2O to grams of H2O
Moles of H2O produced = 32 g C3H8 x 1 mole/44 g x 4 moles H2O/mole C3H8 = 2.909 moles H2O
Grams H2O produced = 2.909 moles H2O x 18 g/mole = 52.36 g = 52 g H2O
You’d need to mix over have the mass of the 450 gram solution
