Answer:
x² = mutiphy by them self
Explanation:
Answer:
In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as
c=4.18Jg∘C
Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.
Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.
In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.
What if you wanted to increase the temperature of 1 g of water by 2∘C ?
This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.
And there you have it. The equation that describes all this will thus be
q=m⋅c⋅ΔT , where
q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature
In your case, you will have
q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C
q=10,450 J
Chechnya. 345. Gcbjshjkfs
Answer:
The total change in enthalpy for the reaction is - 81533.6 J/mol
Explanation:
Given the data in the question;
Reaction;
HCl + NaOH → NaCl + H₂O
Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J
Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
so, 0.0500 moles of H₂O produced
Volume of solution = 50.mL + 50.mL = 100.0 mL
Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g
now ,
Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal
we know that; The specific heat of water(H₂O) is 4.18 J/g°C.
so we substitute
q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28
q_soln = 2842.4 + 1234.28
q_soln = 4076.68 J
Enthalpy change for the neutralization is ΔH
ΔH = -q_soln / mole of water produced
so we substitute
ΔH = -( 4076.68 J ) / 0.0500 mol
ΔH = - 81533.6 J/mol
Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol
