Answer:
The second choice, or flammability.
Explanation:
The flammability of something is how easy it is for it to burn or ignite.
Answer:
1.93
Explanation:
Moles of C
H
COOH = 38/1000 × 0.50 = 0.019mol
Moles of CHCOONa = Mass/Molar mass = 2.64/144.10 = 0.018321mol
Final pH = pKa + log([CHCOONa]/[CHCOOH]
= -log Ka + log(mols of CHCOONa]/mols of CHCOOH
= -log(6.5 × 10^(-5)) + log (0.018321/0.019)=4.17
change in pH = final - initial pH
= 4.17 - 2.24
=1.93
Answer:
The answer to your question is 2.32 atm
Explanation:
Data
P = ?
n = 0.214
V = 2.53 L
T = 61°C
R = 0.082 atm L/mol°K
Formula
PV = nTR
solve for P
P = nRT/V
Process
1.- Calculate the temperature in K
°K = °C + 273
°K = 61 + 273
= 334
2.- Substitution
P = (0.214 x 0.082 x 334) / 2.53
3.- Simplification
P = 5.86/2.53
4.- Result
P = 2.32 atm
