Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
Answer:
Elements
Explanation:
Each kind of atom is an element. Elements are pure substances that cannot be broken down into simpler substances by ordinary chemical means.
OPTION C
<h2>POTENTIAL ENERGY </h2>
THANK U
Answer:
What is the amount of energy needed to change 20g of ice at 0°C to steam at 100°C?
A mattress you can fall back on.
Use the formula
q = m·ΔHf
where q = heat energy; m = mass; ΔHf = heat of fusion
q = (20 g)x(334 J/g)
q = 6680 J
Step 2: Heat required to raise the temperature of 0 °C water to 100 °C water
q = mcΔT
q = (20 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]
q = (20 g)x(4.18 J/g·°C)x(100 °C)
q = 8360 J
Heat required to raise the temperature of 0 °C water to 100 °C water = 8360 J
Heat required to convert 100 °C water to 100 °C steam
q = m·ΔHv
where q = heat energy; m = mass; ΔHv = heat of vaporization
q = (20 g)x(2257 J/g)
q = 45140 J
Heat required to convert 100 °C water to 100 °C steam = 45140 J
Total energy required: 6680+8360+45140=60180 Joules
Or practically understanding it: about 20s in a 3000W kettle.
