<span>43.83g of nacl are required to </span><span>make 500.0 ml of a 1.500 m solution.</span>
Answer: 0.0826mol
PV=nRT
n=PV/RT
n=(1atm)(2.1L)/(310K)(0.082057L*atm/mol*K)=0.0826mol
Explanation:
1)
Mass of NaOH = m
MOlar mass of NaOH = 40 g/mol
Volume of NaOH solution = 1.00 L
Molarity of the solution= 1.00 M
A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.
Upto two significant figures mass should be determined.
2)
(dilution equation)
Molarity of the NaOH solution = 
Volume of the solution = 
Molarity of the NaOH solution after dilution = 
Volume of NaOH solution after dilution= 
A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.
Upto three significant figures volume should be determined.
Answer:
length
Explanation:
cm measures length. Think of a ruler.
Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).
What is standard entropy?
The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."
Calculation:
Balancing the given reaction following-
1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)
ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]
Here S° = standard entropy of the system
Insert into the aforementioned equation all the typical entropy values found in the literature:
ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]
⇒ΔS° = - 99.4 J/K
Therefore, the standard entropy, ΔS° is -99.4 J/K.
Learn more about standard entropy here:
brainly.com/question/14356933
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