Use the drop-down menus to name these structures F FICH H

Butane gas reacts with oxygen gas to give carbon dioxide gas and water vapor (gas). If you mix butane and oxygen in the correct

Fittoniya [83]

Answer:

118.776 mmHg

Explanation:

The equation of the reaction is;

C4H10(g) + 13/2 O2(g) ------> 4CO2(g) + 5H20(g)

Now the mole ratio according to the balanced reaction equation is;

1 : 6.5 : 4 : 5

Hence, the total number of moles present = 1 + 6.5 + 4 + 5 = 16.5 moles

Mole fraction of water vapour = 5/16.5 = 0.303

We also know that;

Partial pressure= mole fraction * total pressure

Partial pressure of H20(g) = 0.303 * 392 mmHg = 118.776 mmHg

7 0

3 years ago

Please help me balance: _NaNO3 + _PbO_ ➡️ Pb(NO3)2 + _Na2O

KengaRu [80]

Answer:

The balanced reaction is given by,

$2NaN(O)3 + PbO$ ⇒ $Pb(NO3)2 + Na2O$

Explanation:

The reaction is as given.

Lets count the number of each elements in the reaction.

In reactant side, number of sodium atoms are 1 , lead are 1, nitrogen are 1 and oxygen are 4.

in product side, number of sodium atoms are 2 , lead are 1 , nitrogen are 2 and oxygen are 7.

So we need to balance sodium and oxygen atoms in the reaction.

There is deficient of sodium and oxygen atoms on reactant side.

Thus, multiply (NaNO3) by 2.

Thus, sodium atoms become 2 , nitrogen 2 and oxygen 6. Total 7 oxygen atoms.

Thus, the balanced reaction is,

$2NaN(O)3 + PbO$ ⇒ $Pb(NO3)2 + Na2O$

7 0

3 years ago

1. By means of orbital diagrams, write down the

Dvinal [7]

Answer:

a) 1s22s22p63s1

b) 3s² 3p¹

c) 1s22s22p63s23p3

d) 3s² 3p⁵

e) 3s2

f) 1s22s22p63s23p2

g) 1s22s22p63s23p4

h) 3s² 3p⁶

3 0

3 years ago

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.