Particles are smaller, a reactant would be exposed to more cobalt atoms. Causing the reaction to happen quicker than with larger particles.
Answer:
Correct option is
B
5 liters of CH
4
(g)NO
2
at STP
No. of molecules=
22.4
5
mol=
22.4
5
×N
A
molecules
A) 5ℊ of H
2
(g)
No. of moles=
2
5
mol=
2
5
×N
A
molecules
B) 5l of CH
4
(g)
No. of moles of CH
4
=
22.4
5
mol=
22.4
5
N
A
molecules
C) 5 mol of O
2
=5N
A
O
2
molecules
D) 5×10
23
molecules of CO
2
(g)
Molecules of 5l NO
2
(g) at STP=5l of CH
4
(g) molecules at STP
Therefore, option B is correct.
Answer : The internal energy change is -2805.8 kJ/mol
Explanation :
First we have to calculate the heat gained by the calorimeter.

where,
q = heat gained = ?
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 fructose = 

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole
Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.
Formula used :

or,

where,
= change in enthalpy = 
= change in internal energy = ?
= change in moles = 0 (from the reaction)
R = gas constant = 8.314 J/mol.K
T = temperature = 
Now put all the given values in the above formula, we get:




Therefore, the internal energy change is -2805.8 kJ/mol
Answer: Empirical formula is 
Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.
<u>Step 1 :</u> Converting each of the given masses into their moles by dividing them by Molar masses.

Molar mass of Carbon = 12.0 g/mol
Molar mass of Hydrogen = 1.0 g/mol
Molar mass of Oxygen = 16.0 g/mol
Moles of Carbon = 
Moles of Hydrogen = 
Moles of Oxygen = 
<u>Step 2: </u>Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value
Smallest number of moles = 13.76 moles



<u>Step 3:</u> Now, the moles ratio of the elements are represented by the subscripts in the empirical formula
Empirical formula becomes = 