List 2 metals that can be used as fuel

Complete the statement<br>qxy- bxy+cxy= xy( )

andrey2020 [161]

Answer:

xy (-b+c+q) is the answer to this

3 0

3 years ago

All of the following are equal to Avogadro's number EXCEPT ____. a. the number of atoms of bromine in 1 mol Br2 b. the number of

Colt1911 [192]

All the following are equal to Avogadro's number EXCEPT a. the number of atoms of bromine in 1 mol Br₂.

1 mol Br₂ contains Avogadro’s number of molecules of Br₂.

However, each molecule contains two atoms of Br, so there are

<em>2 × Avogadro’s number of Br atoms </em>in 1 mol Br₂.

8 0

2 years ago

Balance the following equation in acidic conditions Phases are optional. S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+

jonny [76]

To balance the the chemical reaction, the number of moles per element is balance is both side of the reaction and also the charge in both sides of the reation. to balnce the reaction:

S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+

2S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+ + e

3 0

3 years ago

A voltaic cell is constructed from an Ni2+(aq)−Ni(s)Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s)Ag+(aq)−Ag(s) half-cell. The in

eduard

Answer:

+1.03 V

Explanation:

The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).

The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:

Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V

Ag⁺(aq) + e⁻ → Ag(s) E° = +0.80 V

The value is calculated by a spontaneous reaction, in which the cell with the greater E° is reduced (gain electrons), and the other is oxidized (loses electrons). So, Ag⁺ reduces.

emf = E°reduces - E°oxides

emf = 0.80 - (-0.23)

emf = +1.03 V

8 0

3 years ago

1. If a solution containing 48.99 g of mercury(II) perchlorate is allowed to react completely with a solution containing 8.564 g

Vitek1552 [10]

1. Chemical eqn:

Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d.p) and for Na2S it is 78.0g/mol (1d.p.)

no. of moles of mercury perchlorate= 48.99÷399.5= 0.12263mol(5s.f.)

no. of moles of Na2S= 8.564÷78.0= 0.10979mol ( 5s.f.)

so no. of moles of Hg(ClO4)2/ no. of moles of Na2S= 1/1 according to the eqn

so no. of moles of Hg(ClO4)2 needed if all Na2S is used up is 1/1×0.10979= 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS/ no. of moles of NaClO4= 1/1

no. of moles of HgS produced= 0.10979mols

molar mass of HgS= 232.7g/mol 1d.p.

grams of solid produced= 232.7×0.10979= 25.5g (3s.f.)

2. reactant in excess is Hg(ClO4)2,

no. of excess moles= 0.12263-0.10979= 0.01284mols

grams of excess reactant= 0.01284×399.5= 5.13g (3s.f.)

3 0

3 years ago