All categories

1 year ago

Which of the following rules is applicable for balancing a chemical equation?

Chemistry

1 answer:

Ber [7]1 year ago

7 0

Answer:

A.) Change only the coefficients

Explanation:

An equation is balanced when there is an equal quantity of each type of element on both sides of a reaction. When balancing an equation, the only way to manipulate the amounts of each element is by changing the coefficient values. The coefficients alter the amount of each molecule in the reaction.

The subscripts cannot be altered. If you were to change the subscripts, you would be altering the amount of atoms in a particular molecule.

You might be interested in

Helium decays to form lithium. Which equation correctly describes this decay?

GrogVix [38]

Answer:

⁴He₂ → ⁴Li₃ + ⁰e₋₁.

Explanation:

Helium decays with the release of an electron to form a lithium nucleus.

The number of nucleons is the same on both sides of the equation.

A neutron decays to form a proton with the release of an electron.

The atomic number increases from 2 to 3 while the atomic mass number stays the same at 4 nucleons.

8 0

3 years ago

Read 2 more answers

The picture is above^ someone help me please :)

Ronch [10]

Answer:

idk plz do not report

Explanation:

8 0

3 years ago

A snowflake contains 2.1 10^18 molecules of water. How many moles of water does it contain?

IgorC [24]

Answer:

divide the # of molecules by avogadros number and get 3.48 x 10^-6

Explanation:

6 0

2 years ago

Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu

nydimaria [60]

Answer : The correct option is, +91 kJ/mole

Solution :

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.

$E^0_{[Pb^{2+}/Pb]}=-0.13V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

$E^0_{cell}=E^0_{cathode}-E^0_{anode}$

$E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}$

$E^0_{cell}=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the standard Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole$

Therefore, the standard Gibbs free energy is +91 kJ/mole

6 0

2 years ago

What is earth's crust mainly composed of

____ [38]

Answer:

The crust is made of solid rocks and minerals. Beneath the crust is the mantle, which is also mostly solid rocks and minerals, but punctuated by malleable areas of semi-solid magma. At the center of the Earth is a hot, dense metal core.

Explanation:

5 0

2 years ago

Read 2 more answers