Answer:
7.00
Explanation:
When the solutions are mixed, the HCl dissociates to form the ions H+ and Cl-. The ion H+ will react with the NH3 to form NH4+. The stoichiometry for this is 1 mol of HCl to 1 mol of H+ to 1 mol of Cl-, and 1 mol of H+ to 1 mol of NH3 to 1 mol of NH4+.
First, let's find the number of moles of each one of them, multiplying the concentration by the volume:
nH+ = 0.15 M * 25 mL = 3.75 mmol
nNH3 = 0.52 M * 25 mL = 13 mmol
So, all the H+ is consumed, and the neutralization is completed, thus pH will be the pH of the solvent (water), pH = 7.00.
When sulfur reacts with zinc metal is forms a compound known as zinc sulfide.
<h3>
Name of the elements</h3>
The names of the elements that make up the compound will be determined from the atomic mass of the elements.
<h3>Which element has atomic number 30</h3>
The element that atomic number of 30 is zinc and it has oxidation number of +2.
<h3>Which element has atomic number 16</h3>
The element that atomic number of 16 is sulfur and it has oxidation number of -2.
<h3>Chemical reaction between sulfur and zinc metal</h3>
Zn²⁺ + S²⁻ --------> ZnS
Thus, when sulfur reacts with zinc metal is forms a compound known as zinc sulfide.
Learn more about zinc sulfide here: brainly.com/question/20806552
#SPJ1
Answer:
HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)
Explanation:
Complete question
Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride. What is the net ionic equation
Equilibrium equation between the undissociated acid and the dissociated ions
HF(aq)⇌H+(aq)+F−(aq)
Sodium hydroxide will dissociate aqueous solution to produce sodium cations, Na+, and hydroxide anions, OH−
NaOH(aq)→Na+(aq)+OH−(aq)
Hydroxide anions and the hydrogen cations will neutralize each other to produce water.
H+(aq)+OH−(aq)→H2O(l)
On combining both the equation, we get –
HF(aq)+Na+(aq)+OH−(aq)→Na+(aq)+F−(aq)+H2O(l)
The Final equation is
HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)
Explanation:
Given -
- An organic compound gives H₂ gas with Na
- On treatment with alkaline iodine it gives yellow ppt.
- On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)
To Find -
- Name the compound and write the reaction involved
Now,
Let A be the organic compound.
Then,
- A + Na → + H₂↑
- A + I₂ → CHI₃ (yellow ppt.)
- A + CrO₃ + H⁺ → C₂H₄O
Now,
Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.
- Functional group of aldehyde = —CHO
And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).
Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.
It means,
We know that 1° alcohol on oxidation gives aldehyde.
Here it gives 2 Carbon aldehyde.
It means,
Here 2 Carbon and 1° alcohol is used.
Now,
Its cleared that Compound A is Ethanol.
Reaction Involved -
- CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
- CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
- CH₃CH₂OH + CrO₃ + H⁺ → CH₃CHO
