The structure of
Alanine is shown below,
Except the carbon atom of carbonyl group which is
Sp² Hybridized, all remaining atoms are
Sp³ Hybridized. The hybridization of each element is depicted in a box below,
Answer:- 0.138 M
Solution:- The buffer pH is calculated using Handerson equation:
acts as a weak acid and 
as a base which is pretty conjugate base of the weak acid we have.
The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:
Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.
So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.
Let's plug in the values in the equation:
Taking antilog:
On cross multiply:
[base] = 1.38(0.10)
[base] = 0.138
So, the concentration of the base that is required to make the buffer is 0.138M.
Answer: The questions looks unclear
Explanation: Periodic table is a table that contains elements arranged according to their increasing atomic number.
1. D belongs to group 4
E. Belongs to group 7
B belongs to group 1
A belongs to group 8. A noble gas.
R belongs to group 3. K belongs to group 6 C belongs to group 1. H belongs to group 8
Answer:
<span>23.6
g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from
15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because
it gets used up and only makes 10.7 g of CO2. </span>
Explanation:
1) Balanced chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O
3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol
4) Convert the reactant masses to number of moles, using the formula
number of moles = mass in grams / molar mass
CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
<span />
O₂: 15.6 g / 32.0 g/mol = 0.4875 moles
5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂
Which is what the first part of the answer says.
6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.
Which is what the second part of the answer says.
7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.
Which is what the chosen answer says.
8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.
