Answer:
pH = 3.34
Explanation:
We can calculate the pH using Henderson-Hasselbach's equation:
pH = pKa + log[tex]\frac{[HCOO^-]}{[HCOOH]}[/tex]for HCOOH, pKa = 3.75.
We can calculate [HCOO⁻] and [HCOOH] using the given number of moles for each one and the final volume:
Final Volume = 44.12 mL + 250 mL = 294 mL294 mL / 1000 = 0.294 L[HCOO⁻] = 0.036 mol / 0.294 L = 0.122 M[HCOOH] = 0.091 mol / 0.294 L = 0.310 MThen we proceed to calculate the pH:
pH = 3.75 + log[tex]\frac{0.122}{0.310}[/tex]pH = 3.34The pH of the resulting solution is 3.34.
Calculation of the pH of the solution:Since
Final Volume = 44.12 mL + 250 mL = 294 mL
Now
= 294 mL / 1000
= 0.294 L
Now
[HCOO⁻] = 0.036 mol / 0.294 L = 0.122 M
[HCOOH] = 0.091 mol / 0.294 L = 0.310 M
So, the pH should be
= 3.75 + log0.122/0.310
= 3.34
hence, The pH of the resulting solution is 3.34.
learn more about reaction here: https://brainly.com/question/20874611
Which one is not an ore of copper
1 Azurite
2 Malachite
3 Haematite
4 Chalcopyrite
Hydrochloric acid reacts with sodium hydroxide to form water and sodium chloride. Hydrochloric acid is an extremely acidic, clear, corrosive liquid. Sodium hydroxide is a very basic white solid.
What can be known about the reactants of this reaction?
They will be clear.
They will not have the properties of sodium hydroxide or hydrochloric acid.
They will be corrosive.
They will have the properties of sodium hydroxide or hydrochloric acid.
The answer is C.
You're welcome!
Which answer would represent 0.001 moles?
Answer:
Which answer would represent 0.001 moles?
A sealed 1.0L flask is filled with 0.500 mols of I_2 and 0.500 mols of Br_2. When the container achieves equilibrium the equilibrium constant is 1.10x10^{-2}. What is the equilibrium concentration of the product, IBr?
Answer:
[IBr] = 0.049 M.
Explanation:
Hello there!
In this case, according to the balanced chemical reaction:
[tex]I_2+Br_2\rightarrow 2IBr[/tex]
It is possible to set up the following equilibrium expression:
[tex]K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110[/tex]
Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of [tex]x[/tex] (reaction extent) would be:
[tex]0.0110=\frac{(2x)^2}{(0.50-x)^2}[/tex]
Which can be solved for [tex]x[/tex] to obtain two possible results:
[tex]x_1=-0.0277M\\\\x_2=0.0245M[/tex]
Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:
[tex][IBr]=2x=2*0.0249M=0.049M[/tex]
Regards!
Set of degenerate orbital in germanium
Answer:
Which set of orbitals are degenerate?
Degenerate is used in quantum mechanics to mean 'of equal energy. ' It usually refers to electron energy levels or sublevels. For example, orbitals in the 2p sublevel are degenerate - in other words the 2px, 2py, and 2pz orbitals are equal in energy, as shown in the diagram.
Explanation:
Answer:
Electron orbitals having the same energy levels are called degenerate orbitals, For example, orbitals in the 2p sublevel are degenerate - in other words the 2px, 2py, and 2pz orbitals are equal in energy.
How does air flow when a low-pressure center occurs in the
atmosphere?
Answer:
A
Explanation:
It is so because the low air pressure create vacuum and the air from high pressure area move toward the low air pressure.
When a sample of gas was placed in a sealed container with a volume of 3.35 L and heated to
75°C, the gas vaporized and the resulting pressure inside the container was 17atm. How
many moles of the gas were present?. Single line text.
Answer:
Explanation:What is the demand factor for three commercial ranges?
Hydrogen peroxide with a concentration of 3.0 percent (3.0 g of H2O2 in 100 mL of solution) is sold in drugstores for use as an antiseptic. For a 10.0-mL 3.0 percent H2O2 solution, calculate (a) the oxygen gas produced (in liters) at STP when the compound undergoes complete decomposition and (b) the ratio of the volume of O2 collected to the initial volume of the H2O2 solution.
Answer:
a) 0.099 L
b) 9.9
Explanation:
Now, given the equation for the decomposition of H2O2;
2H2O2(l) ------> 2H2O(l) + O2(g)
Mass of H2O2;
percent w/v concentration = mass/volume * 100
volume = 10.0-mL
percent w/v concentration = 3.0 percent
mass of H2O2 = x
3 = x/ 10 * 100
30 = 100x
x = 30/100
x = 0.3 g of H2O2
Number of moles in 0.3 g of H2O2 = mass/ molar mass
Molar mass of H2O2 = 34.0147 g/mol
Number of moles in 0.3 g of H2O2 = 0.3g/34.0147 g/mol
= 0.0088 moles
From the reaction equation;
2 moles of H2O2 yields 1 mole of oxygen
0.0088 moles of H2O2 = 0.0088 * 1/2 = 0.0044 moles of oxygen
If 1 mole of oxygen occupies 22.4 L
0.0044 moles of oxygen occupies 0.0044 * 22.4/1
= 0.099 L
b) initial volume of the H2O2 solution = 10 * 10-3 L
Hence, ratio of the volume of O2 collected to the initial volume of the H2O2 solution = 0.099 L/10 * 10-3 L = 9.9