1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Valentin [98]
3 years ago
8

Water acts as a solvent. a. True b. False

Chemistry
1 answer:
Talja [164]3 years ago
7 0
True. Water acts as a solvent as the solute dissolves into water. 
You might be interested in
What time of the year can you see Scorpius
enyata [817]

Answer:

In the northern hemisphere I believe, Scorpius is most visible by looking to the south during July and August around 10:00 PM. The constellation remains visible until mid-September. In the southern hemisphere, Scorpio appears very high in the northern part of the sky until close to the end of September.

8 0
3 years ago
Read 2 more answers
When a 14.2-g sample of mercury (II) oxide is decomposed into its elements by heating, 13.2 g Hg is obtained. What is the percen
nevsk [136]
Mercury (ii) oxide is made up of mercury and oxygen. The total mass of mercury (ii) oxide is 14.2 g, after decomposition 13.2 g of mercury were formed, therefore the mass of oxygen 1 g (14.2 g -13.2 g).
Percentage of oxygen = (1/14.2)×100 = 7.04%
Percentage of mercury = (13.2/14.2) × 100 = 92.96%
Therefore, percentage composition of the compound, oxygen is 7.04% and mercury is 92.96%.
4 0
3 years ago
An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6
ExtremeBDS [4]

Answer :

(a) The concentration of Pb^{2+} is, 0.0337 M

(b) The concentration of Pb^{2+} is, 6.093\times 10^{32}M

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Pb\rightarrow Pb^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

The balanced cell reaction will be,  

Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}

E^o=0.34V-(-0.13V)=0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}

[Pb^{2+}]=0.0337M

Therefore, the concentration of Pb^{2+} is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction:  Pb^{2+}+2e^-\rightarrow Pb

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}

E^o=-0.13V-(0.34V)=-0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}

[Pb^{2+}]=6.093\times 10^{32}M

Therefore, the concentration of Pb^{2+} is, 6.093\times 10^{32}M

6 0
3 years ago
Based on the article "Will the real atomic model please stand up?,” describe one major change that occurred in the development o
nydimaria [60]
I don't know this article, but I do know some major changes: first, the change from the plum pudding model (no nucleus, just electrons) to the gold foil experiment, which had Rutherford shoot alpha particles at a sheet of gold only to find them rebounding, proving the existence of a positively charged mass, i.e a nucleus, in the atom. However, this changed again when Bohr realized that the negatively charged electrons should be attracted to the positively charged center, so that there must be something else inside the nucleus. 
3 0
2 years ago
Read 2 more answers
Which element has the ground state electron configuration 1s22s22p63s23p3?
Natalka [10]
Its phosphorus (P)In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.
4 0
3 years ago
Other questions:
  • The stomach and intestines allow a person to grow and survive by helping the body
    6·1 answer
  • What are the answers of the parts? All of them please.
    9·1 answer
  • Please help I will reward brainly THERE ARE 3 ANSWERS
    15·2 answers
  • You are on an alien planet where the names for substances and the units of measures are very unfamiliar.
    12·1 answer
  • Calculate the percentage of a solution formed by dissolving 10g of glucose in 240g of water
    13·1 answer
  • Douglasite is a mineral with the formula 2KCl • FeCl2 • 2H2O. Calculate the mass percent of douglasite in a 455.0-mg sample
    11·1 answer
  • A(n) _______ solution can dissolve more solute. A. Unsaturated B. Saturated
    6·1 answer
  • Give the formula of a reagent which will form a
    9·1 answer
  • Problem
    14·1 answer
  • Calculate mL (4 sf) of 0.7500 M sodium hydroxide required to neutralize 35.00 mL of 0.7500 M phosphoric acid. Please input numbe
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!