I will, I am in a chemistry class right now. What’s up?
Answer:
The Enthalpy of neutralization
Explanation:
The reaction of a base (KOH) with an acid (HCl) produce water and its salt (KCl) is called <em>Neutralization Reaction.</em>
This neutralization releases 57kJ/mol.
As the type of enthalpy is due the type of reaction. This enthalpy is:
<h3>The Enthalpy of neutralization</h3>
Answer:
D
Explanation:
I think it's D but not 100% sure
Answer:
0.501 M
Explanation:
First we <u>convert 1.513 g of KHP into moles</u>, using its <em>molar mass</em>:
- 1.513 g ÷ 204.22 g/mol = 7.41x10⁻³ mol = 7.41 mmol
As <em>1 mol of KHP reacts with 1 mol of NaOH</em>, in 14.8 mL of the NaOH solution there were 7.41 mmoles of NaOH.
With the above information in mind we can <u>calculate the molarity of the NaOH solution</u>:
- 7.41 mmol / 14.8 mL = 0.501 M
Answer:
200.6cm³
Explanation:
Using Charles law equation as follows:
V1/T1 = V2/T2
Where;
V1 = initial volume (cm³)
V2 = final volume (cm³)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the question;
initial volume (V1) = 254cm³
final volume (V2) = ?
Initial temperature (T1) = 72.6°C = 72.6 + 273 = 345.6K
Final temperature (T2) = 273K
V1/T1 = V2/T2
254/345.6 = V2/273
Cross multiply
345.6 × V2 = 273 × 254
345.6V2 = 69342
V2 = 69342 ÷ 345.6
V2 = 200.6cm³
