We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH
The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])
We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87
So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.
Answer:
cinnamic acid - 150 mg
cis-stilbene - 100 μL
trans- stilbene - 100 mg
pyridinium tribromide - 200-385 mg
For this data:
moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols
Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g
cis-stilbene (100 ul = 0.1 ml)
moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols
Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g
trans-stilbene
moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols
Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g
Explanation:
Answer:
H₂S(aq)+ 2 LiOH(aq) → Li₂S(aq) + 2 H₂O(l)
6 HI(aq) + 2 Al(s) → 2 AlI₃(aq) + 3 H₂(g)
2 H₂SO₄(aq) + TiO₂(s) → Ti(SO₄)₂(aq) + 2 H₂O(l)
H₂CO₃(aq) + 2 LiOH(aq) → Li₂CO₃(aq) + 2 H₂O(l)
Explanation:
H₂S(aq)+ 2 LiOH(aq) → Li₂S(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
6 HI(aq) + 2 Al(s) → 2 AlI₃(aq) + 3 H₂(g)
This is a single displacement reaction.
2 H₂SO₄(aq) + TiO₂(s) → Ti(SO₄)₂(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
H₂CO₃(aq) + 2 LiOH(aq) → Li₂CO₃(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
Answer:
The absorption and strength of the H-beta lines change with the temperature of the stellar surface, and because of this, one can find the temperature of the star from their absorption lines and strength. To better comprehend, let us look into the concept of the atom's atomic structure.
Atoms possess distinct energy levels and these levels of energy are constant, that is, the temperature has no influence on it. However, temperature possesses an influence on the electron numbers found within these levels of energy. Therefore, to generate an absorption line of hydrogen in the electromagnetic spectrum's visible band, the electrons are required to be present in the second energy level, that is when it captivates a photon.
Therefore, after captivating the photons the electrons jump from level 2 to level 4, which shows that there is an increase in the stellar surface temperature and at the same time one can witness a decline in the strength of the H-beta lines. In case, if the temperature of the surface increases too much, then one will witness no attachment of electron with the hydrogen atom and thus no H lines, and if the temperature of the surface becomes too low, then the electrons will stay in the ground state and no formation of H lines will take place in that condition too.
Hence, to generate a very robust H line, after captivating photons the majority of the electrons are required to stay in the second energy level.
Basically any animal group that doesnt have any backbones. for example fish birds different types of mammals and reptiles.
hope this helps!
