Answer:
ΔS Rx = -120, 5 J/K
Explanation:
The ΔS in a reaction is defined thus:
ΔS Rx = ∑ n S°products - ∑ m S°reactants
For the reaction:
N₂(g) + 2 O₂(g) → 2NO₂(g)
ΔS Rx = 2 mol × 240,5 
- [ 1 mol × 191,5 + 2 mol × 205,0 ]=
<em>ΔS Rx = -120, 5 J/K</em>
A negative value in ΔS means a negative entropy of the process. Doing this process entropycally unfavorable.
I hope it helps!
Answer:
32.25 J
Explanation:
Heat, H = ?
Mass, m = 5g = 5 / 1000 = 0.005 Kg
Initial Temperature = 75.0° C
Final Temperature = 25.0° C
Temperature change, ΔT = Final - Initial = 50° C
Specific heat capacity of lead, C = 129 (J/kg°C)
The relationship between these quantities is given by the equation
H = mCΔT
H = 0.005 * 129 * 50
H = 32.25 J
The element itself is Yttrium, which is a transition metal and not that reactive. Since it has 39 protons it will also have 39 electrons.
The missing components of the neutralisation reaction include the following:
- KBr
- KBr 2NH4OH
- KBr 2NH4OH2HNO2
<h3>What is neutralisation reaction?</h3>
Neutralisation reaction is defined as the type of reaction that leads to the formation of salt and water when an acid and a base reacts.
From the reactions given the missing components are replaced as follows:
- H2SO4 + 2NH4OH --> (NH4)2SO4 + 2H2O
- 2HNO3 + Mg(OH)2 --> Mg(NO3)2 + 2H2O
Learn more about acids here:
brainly.com/question/26353151
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Answer:
We need 7.5 mL of the 1M stock of NaCl
Explanation:
Data given:
Stock = 1M this means 1 mol/ L
A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL
Step 2: Calculate the volume of stock we need
The moles of solute will be constant
and n = M*V
M1*V1 = M2*V2
⇒ with M1 = the initial molair concentration = 1M
⇒ with V1 = the volume we need of the stock
⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L
⇒ with M2 = the concentration of the new solution = 0.15 M
1*V1 = 0.15*(50)
V1 = 7.5 mL
Since 0.0075 L of 1M solution contains 0.0075 moles
50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M
We need 7.5 mL of the 1M stock of NaCl
