CHEMISTRY HIGH SCHOOL

To prevent accidental poisoning, you should NOT __________.a. mix chemicals in well ventilated areas
b. follow the instructions on the label
c. keep materials in unlabeled containers
d. use chemicals in well ventilated areas

Answers

Answer 1
Answer: Answer: option C. keep materials in unlabeled containers.


Justification:


Indeed, keeping materials in unlabeled containers is a risky, dangerous, bad practice.


Labels should be according to international standards. That is, further to the name of the product, labels must signal the danger associated with the product: whether it is poison, toxic, flammable, acid, caustic, explosive, or harmful to the body or the enviroment.


On the other hand, the other options, a. mix chemicals in well ventilated areas, b. follow the instructions on the label, and d. use chemicals in well ventilated areas, are good safety practices, among many others.
Answer 2
Answer:

Answer: C

Explanation:

Keep materials in unlabeled containers

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What property of an atom determines its general reactivity?
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In which reaction is soap a product?(1) addition (3) saponification
(2) substitution (4) polymerization

Answers

Saponification because you can rearrange the letters to spell soap.

Which explains why carbon is a component of so many elements?A. It appears in more than one form
B. It is the most abundant element
C. It can react with both metals and nonmetals
D. It can form stable compounds with other elements

Answers

Carbon is a component of so many elements because It can form stable compounds with other elementsand also bond with them.

A certain compound is made up of one phosphorus (P) atom, three chlorine (Cl) atoms, and one oxygen (O) atom. What is the chemical formula of this compound?

Answers

Ok so we know that all of these are non metals therefore they will bond using covalent bonds.
Phoshorus can have 3 bonds Chlorine can have 1 bond and Oxygen can two bonds.
Cl-P=O
The chemical formula I belive is PClO hope this helps :).

What does Nb stand for on the periodic table? Pls help! Need aware soon

Answers

it stands for niobium which was discovered in 1801 by Charles Hatchett of Brittan. It is ussed in welding rods, air fraim systems, superconductive magnets. It gets its name from iobe, the daughter of King Tantalus of the Greek myth. Niobium was considered identical to Tantalum, named after Tantalus, until 1884. Resists corrosin due to an oxide film. It is used in aviation industry
it stands for niobium 

Which expression could represent theconcentration of a solution?
(1) 3.5 g
(2) 3.5 M
(3) 3.5 mL
(4) 3.5 mol

Answers

There are a couple of ways in which you can express the concentration of a solution, and here they are: gram per liter (g/L), molarity (M), parts per million (ppm.), and percents (%). 
As you can see, only M appears in your answers, which means that the correct option should be (2) 3.5 M.

The expression for representation of concentration of a solution has been 3.5 M. Thus, option 2 is correct.

Concentration has been defined as the amount of substance present in a volume of solution. The concentration has been given by varying units. The concentration has been given by mass per unit volume.

It can be represented by moles/L or g/L. The expression that has been similar to moles/L has been molarity.

Thus, the expression for representation of concentration of a solution has been 3.5 M. Thus, option 2 is correct.

For more information about concentration, refer to the link:

brainly.com/question/3045247

Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions. Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) S (s) (nonspontaneous)? Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq) (nonspontaneous)? 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq) (spontaneous?) Are these correct?

Answers

A electrochemical reaction is said to be spontaneous, if E^(0) cell is positive. 

Answer 1:
Consider reaction: Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s) 

The cell representation of above reaction is given by;
    S^(2-)/S // Ni^(2+)/Ni

Hence, E^(0)cell = E^(0) Ni^(2+/Ni) - E^(0) S/S^(2-)
we know that, {E^(0) Ni^(2+)/Ni = -0.25 v
and {E^(0) S/ S^(2-) = -0.47 v

Therefore, E^(0) cell = - 0.25 - (-0.47) = 0.22 v

Since,  E^(0) cell is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2: 
Consider reaction: Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)

The cell representation of above reaction is given by;
    H_(2) / H^(+) // Pb^(2+) /Pb

Hence, E^(0)cell = E^(0) Pb/Pb^(2+) - E^(0) H_(2)/H^(+)
we know that, {E^(0) Pb^(2+)/Pb = -0.126 v
and {E^(0) H_(2)/ H^(+) = -0 v

Therefore, E^(0) cell = - 0.126 - 0 = -0.126 v

Since,  E^(0) cell is negative, hence cell reaction is non-spontaneous.

....................................................................................................................

Answer 3: 
Consider reaction: 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)

The cell representation of above reaction is given by;
    Cr/Cr^(2+) // Ag^(+)/Ag

Hence, E^(0)cell = E^(0) Ag^(+)/Ag - E^(0) Cr/Cr^(2+)
we know that, {E^(0) Ag^(+)/Ag = -0.22 v
and {E^(0) Cr/ Cr^(2+) = -0.913 v

Therefore, E^(0) cell = - 0.22 - (-0.913) = 0.693 v

Since,  E^(0) cell is positive, hence cell reaction is spontaneous

Answer: Ni^(2+)(aq)+S^(2-)(aq)\rightarrow Ni(s)+S(s)  : non spontaneous

Pb^(2+)(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)  : non spontaneous

2Ag^(+)(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^(2+)(aq)  : spontaneous

Explanation:

a) Ni^(2+)(aq)+S^(2-)(aq)\rightarrow Ni(s)+S(s)

Here S undergoes oxidation by loss of electrons, thus act as anode. Ni undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_(cathode)- E^0_(anode)

Where both E^0 are standard reduction potentials.

E^0_([Ni^(2+)/Ni])=-0.25V

E^0_([S^(2-)/S])=0.407VV

E^0=E^0_([Ni^(2+)/Ni])- E^0_([S^(2-)/S])

E^0=-0.25-(0.407V)-0.657V

As value of E^0 is negative, the reaction is non spontaneous.

b)Pb^(2+)(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)

Here Hydrogen undergoes oxidation by loss of electrons, thus act as anode. Pb undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_(cathode)- E^0_(anode)

Where both E^0 are standard reduction potentials.

E^0_([Pb^(2+)/Pb])=-0.13

E^0_([H^(+)/H_2])=0V

E^0=E^0_([Pb^(2+)/Pb])- E^0_([H^(+)/H_2])

E^0=-0.13-(0V)=-0.13V

As value of E^0 is negative, the reaction is non spontaneous.

c) 2Ag^(+)(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^(2+)(aq)

Here Cr undergoes oxidation by loss of electrons, thus act as anode. Ag undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_(cathode)- E^0_(anode)

Where both E^0 are standard reduction potentials.

E^0_([Ag^(+)/Ag])=+0.80V

E^0_([Cr^(2+)/Cr])=-0.913V

E^0=E^0_([Ag^(+)/Ag])- E^0_([Cr^(2+)/Cr])

E^0=+0.80-(-0.913V)=1.713V

As value of E^0 is positive, the reaction is spontaneous.
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