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charle [14.2K]
3 years ago
5

What is the pH of a 10% HCL Solution?

Chemistry
2 answers:
Sholpan [36]3 years ago
8 0
The solution is 11
 Hope this helps
omeli [17]3 years ago
4 0
The PH of 10% HCL solution is 11
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Explain how the presence of several oxygen atoms ina compound containing an-OH group can make the compound acidic.
Lina20 [59]
<span>In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X"
</span>

The factors to consider

Two of the factors which influence the ionisation of an acid are:

<span>the strength of the bond being broken,the stability of the ions being formed.</span>

In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar.

7 0
3 years ago
According to the following position vs. Time graph the bicyclist was
julia-pushkina [17]

Answer:

According to the following position vs. Time graph the bicyclist was : D

D. Not moving for the first two seconds, then begin moving at the constant speed.​

Explanation:

Speed = \frac{Distance}{time}

Along Y-axis = Distance in meters

Along X-axis = Time taken in second

  • For first 2 seconds the The object has not changed its position.There is Zero distance covered.So , Speed = 0
  • After 2 second,The Object changed the distance by equal amount in equal interval of time.(Uniform Speed).Hence at each point after 2 second, there is same value speed
3 0
3 years ago
Where are elements with similar chemical properties found on the periodic table?
LekaFEV [45]
All the members of a family of elements have the same number of valence electrons and similar chemical properties. The horizontal rows on the periodic table are called periods.
8 0
3 years ago
Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
Total number of atoms in 4so2​
Sunny_sXe [5.5K]

Answer:

10

s=8

o=2

Hope it helps to

3 0
3 years ago
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