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serious [3.7K]
2 years ago
13

??? Ipc helppp pllzz

Chemistry
1 answer:
Umnica [9.8K]2 years ago
5 0
The correct answers are
-formation of a precipitate
-bubble formation
-color change
-temperature change
-odor formation

The only one that isn’t correct is change in state of matter. A change in a state of matter does not mean it’s a chemical change. For example, water boiling so it turns into gas is not a chemical change, and is a physical one. Also, water can freeze and turn into ice, which is also still a physical change. If something changes state of matter, it does not necessarily mean it’s a chemical change.
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What is the number of moles in 15.0 g AsH3?
pentagon [3]

We are given with the mass of Arsine (AsH_{3}

The mass of arsine is 15g

there is a relation between moles, mass and molar mass of any compound which is

moles=\frac{mass}{molarmass}

The molar mass of Arsine = atomic mass of As + 3X atomic mass of H

the molar mass of Arsine = 74.92 + 3X 1 = 77.92 g/mol

Let us calculate the moles as

moles=\frac{15}{77.92}=0.192mol


7 0
3 years ago
What is the PH of a 7.5x10^-3 M OH solution
Triss [41]
pH: 11.8750612634
pOH: 2.12493873661
[H+]: 1.33333333333E-12
[OH-]: 0.0075
BASE
5 0
3 years ago
Read 2 more answers
Determine the molality of an aqueous solution that is 16.5 percent urea by mass
yarga [219]

Take a hypothetical sample of exactly 100 grams of the solution.  

(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea  

((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O  

(0.2664  mol) /0.0840  (kg) = 3.17143mol/kg = 3.18m urea


7 0
3 years ago
Please help with 14-16 <br> Thank you!
Natalka [10]
I don't know about 14, but 15 is (4), because a liquid draws in heat to turn into a gas. 16 is (2), because to turn into a cold solid, something has to release heat.
4 0
3 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
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