Why do you think that the electrons are not shown?

How many hours are in 1987 seconds. Please explain

blagie [28]

Just divide!

60 seconds = 1 min

60 mins = 1 hour

1987 seconds divided by 60 seconds = 33.11 minutes

Your answer would be 33 minutes and 11 seconds

8 0

3 years ago

Ri

trasher [3.6K]

Answer:

Ammonia is a compound.

Explanation:

A compound is composed of two or more separate elements. Ammonia is NH3, which happens to be two different elements.

6 0

4 years ago

Tetrachloromethane,CC15,is classified as a

prohojiy [21]

Correct Answer: compound because the atoms of the elements are combined in a fixed proportion.

7 0

4 years ago

Which of the following represents an element? со H2CO3 02 ОН-

LenKa [72]

Answer:

Co (assuming there's a typo in not capitalizing the first letter)

Explanation:

The elements are depicted in the periodic table. Those are fundamental atoms that are not bonded to anything else but exist in their natural states without forming a bond with anything. That said, we're interested in a monoatomic species listed (the species which consist from only one single atom).

Out of all the choices, we have Co, cobalt, which is a single atom and can be found as an element with the atomic number 27.

Carbonic acid, $H_2CO_3$ , consists of 3 different elements: hydrogen (atomic number of 1), carbon (atomic number of 6) and oxygen (atomic number of 8). It's a compound rather than an element, as it consists of several atoms bonded together.

Oxygen, $O_2$ , is also a molecule. It consists of more than a single atom, even though the two atoms are identical. We have a compound, oxygen gas, which consists of two same oxygen (O) atoms (or elements).

Hydroxide anion, $OH^{-}$ , also consists of two elements: oxygen and hydrogen. It also contains a charge on it, so rather than being a molecule, this is now an ion. Since it consists of more than a single atom, this is not an element as well.

6 0

4 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

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

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

4 years ago