Explanation:
The given data is as follows.
, 
= 2257 kJ/kg, 
= ?
For water, 
= 4.184 
Formula to calculate heat of vaporization is as follows.
= 
Hence, putting the values into the above formula as follows.
=
= 
= 2257 kJ/kg - 376.56 kJ/kg
= 1880.44 kJ/kg
Thus, we can conclude that enthalpy of liquid water at 
is 1880.44 kJ/kg.
Answer:
6.7 x 10²⁶molecules
Explanation:
Given parameters
Mass of CO₂ = 4.9kg = 4900g
Unknown:
Number of molecules = ?
Solution:
To find the number of molecules, we need to find the number of moles first.
Number of moles = 
Molar mass of CO₂ = 12 + 2(16) = 44g/mol
Number of moles = 
= 111.36mole
A mole of substance is the quantity of substance that contains the avogadro's number of particles.
1 mole = 6.02 x 10²³molecules
111.36 moles = 111.36 x 6.02 x 10²³molecules = 6.7 x 10²⁶molecules
I think so... I'm currently learning this too but you should be correct
From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ
<h3>What is Hess' law?</h3>
Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.
From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:
A---> C = A ---> B + B ---> C
ΔH of A---> C = 30 kJ + 60 kJ
ΔH = 90 kJ
Therefore, the enthalpy of the reaction A---> C is +90 kJ
The above reaction A---> C can be shown in the enthalpy diagram below:
A -------------------> C (ΔH = +90 kJ)
\ /
\ / (ΔH = +60 kJ)
(ΔH = +30 J) \ /
> B
Learn more about enthalpy and Hess law at: brainly.com/question/9328637
Scientists use the physical and chemical properties to help them identify and classify matter. These physical and chemical properties are in a macro-perspective, in which these matter contains compounds, elements and atoms. Hence, matter can be classified in various ways, <span><span>
1. </span>Atomic number either atomic mass each element has</span>
<span><span>2. </span>By substance of that matter either pure substance or mixed substance</span> <span>
3. If they cannot reduce a certain substance into a much smaller quantified atomic structure then they they’ll use (2) to identify and classify it.</span>
