Find the mass of 0.456 mol of Ca(OH)2 Please if you can show the work.
2 answers:
<h2>Answer:</h2>
<em>Hey, </em>
<h3><u>QUESTION)</u></h3>A) Let's first calculate the molar mass M of the molecule :
<em>M(Ca(OH)₂) = M(Ca) + 2M(O) + 2M(H) </em>
- M(Ca(OH)₂) = 40 + 2 x 16 + 2 x 1
- M(Ca(OH)₂) = 74 g/mol
B) Let's deduct the corresponding mass m :
<em>m = M x n </em>
- m = 74 x 0.456
- m = 33.744 g
- m ≈ 34,0 g
The mass of calcium hydroxide corresponding to this quantity of substance (in mol) is therefore 34 g.
You might be interested in
Complete Question
The complete question is shown on the first question
Answer:
a) The duty of the heat exchanger is given as 6.8658 KJ /sec
b) The temperature of the water leaving the exchanger is TOUT = 29.84 ⁰C
c) The log mean difference is given as TZ = 47.317 ⁰ C
d) the UA value is UA = 145.10
Explanation:
The explanation is uploaded on the first and second ,third and fourth image
The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
<h3>What is Enthalpy of Vaporization ?</h3>The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.
<h3>How to find the energy change from enthalpy of vaporization ?</h3>To calculate the energy use this expression:
where,
Q = Energy change
n = number of moles
= Molar enthalpy of vaporization
Now find the number of moles
Number of moles (n) =
=
= 0.5 mol
Now put the values in above formula we get
[Negative sign is used because Br₂ condensed here]
= - (0.5 mol × 15.4 kJ/mol)
= - 7.7 kJ
Thus from the above conclusion we can say that The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
Learn more about the Enthalpy of Vaporization here: brainly.com/question/13776849
#SPJ1