Answer:
When the operation of the voltaic cell, which is formed of an aluminum and silver strip takes place, the atom of aluminum loses three of its electrons and the Al3+ formed moves within the solution. The Al3+ ion gets dissolved within the solution and the electrons lost in the process moves through the wire and get acquired by the ions of silver, which then get reduced to solid Ag resulting in the mass gain of silver strip.
N2 + 3H2 ---> 2NH3
mass of N2 = 28g
mass of H2 = 2g
mass of NH3 = 17g
according to the reaction:
28g N2----------------- 3*2g H2
85,5g N2-------------------- x
x = 18,32g H2 >>> so, nitrogen is excess
according to the reaction:
2*3g H2---------------------- 2*17g NH3
17,3g H2 ------------------------- x
x = 98,03g NH3
<u>answer: 98,03g of NH3</u>
Answer : The enthalpy change during the reaction is -6.48 kJ/mole
Explanation :
First we have to calculate the heat gained by the reaction.

where,
q = heat gained = ?
m = mass of water = 100 g
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:


Now we have to calculate the enthalpy change during the reaction.

where,
= enthalpy change = ?
q = heat gained = 23.4 kJ
n = number of moles barium chloride = 

Therefore, the enthalpy change during the reaction is -6.48 kJ/mole
Answer:6.719Litres of Cl2 gas.
Explanation:According to eqn of rxn
2Na +Cl2=2NaCl
P=689torr=689/760=0.91atm
T=39°C+273=312K
according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl
But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW
MW of NaCl=23+35.5=58.5g/mol
n=28g(mass given of NaCl)/58.5
n=0.479moles of NaCl
Going back to the reaction,
if 1moles of Cl2 produces 2moles of NaCl
x moles of Cl2 will give 0.479moles of NaCl.
x=0.479*1/2
x=0.239moles of Cl2.
To find the volume, we use ideal ggas eqn,PV=nRT
V=nRT/P
V=0.239*0.082*312/0.91
V=6.719Litres
<span>Answer Choices:
A) Ca
B) O
C) Cl
D) s</span>
<span>the formula is Li2X, so the charge on the X anion must be 2-
the ion is X2- elements in group 6A form monatomic ions with a 2- charge. In your list that is O or Po
If the element can accomodate 12 electrons then it can have an expanded octet. Only elements in period 3 and higher can have expanded octets.
So you are looking for a group 6A element in period 3 or higher.
cA would not intereact with LI BECAUSE ITS A METAL The only element that fits the bill is D) S</span>