Answer : The enthalpy change for the process is 52.5 kJ/mole.
Explanation :
Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution
![q=[q_1+q_2]](https://tex.z-dn.net/?f=q%3D%5Bq_1%2Bq_2%5D)
![q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]](https://tex.z-dn.net/?f=q%3D%5Bc_1%5Ctimes%20%5CDelta%20T%2Bm_2%5Ctimes%20c_2%5Ctimes%20%5CDelta%20T%5D)
where,
q = heat released by the reaction
= heat absorbed by the calorimeter
= heat absorbed by the solution
= specific heat of calorimeter = 
= specific heat of water = 
= mass of water or solution = 
= change in temperature = 
Now put all the given values in the above formula, we get:
![q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]](https://tex.z-dn.net/?f=q%3D%5B%2812.1J%2F%5EoC%5Ctimes%206.1%5EoC%29%2B%28100.0g%5Ctimes%204.18J%2Fg%5EoC%5Ctimes%206.1%5EoC%29%5D)
Now we have to calculate the enthalpy change for the process.
where,
= enthalpy change = ?
q = heat released = 2626.61 J
n = number of moles of copper sulfate used = 
Therefore, the enthalpy change for the process is 52.5 kJ/mole.
Answer:
B
Explanation:
Two quick google searches later I grace you with the google has bestowed upon me
start the balancing by writing down how many atoms there are per element. we’ll use this as an example:
C3H8 + O2 --> H2O + CO2
C = 3 C = 1
H = 8 H = 2
O = 2 O = 3
balance the carbon first, as it is easiest to do. add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation:
C3H8 + O2 --> H2O + (3)CO2
now there are 3 carbon atoms on each side. however, when you do this, you multiply the amount of oxygen atoms you had. therefore, now, there are 6 carbon atoms in 3CO2, plus that other oxygen atom in H2O. you now have 7 O atoms instead of 3.
C = 3 C = 3
H = 8 H = 2
O = 2 O = 7
now let’s move on to the hydrogen atoms.
C3H8 + O2 --> H2O + 3CO2
you have 8 hydrogen atoms on the left side, and 2 on the right. in order to balance them, you have to multiply the right side’s hydrogen atoms by 4. 4(2) = 8.
C3H8 + O2 --> (4)H2O + 3CO2
now both hydrogen and carbon atoms are balanced. same amount on both sides. however, your oxygen atoms have changed due to the multiplying (right side). you now have 10 of them.
C = 3 C = 3
H = 8 H = 8
O = 2 O = 10
now we balance the oxygen atoms. multiply the left side of the equation’s oxygen atoms by 5. 5(2) = 10
C3H8 + (5)O2 --> 4H2O + 3CO2
the chemical equation is all balanced. basically, just multiply with numbers until it equals the same amount on both sides.
C = 3 C = 3
H = 8 H = 8
O = 10 O = 10
