Answer:
d) 0.1202 M
Explanation:
Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.
NaOH + HA → NaA + H₂O
The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:
22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol
The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.
The molar concentration of HA is:
2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M
Answer:
The partial pressure of the other gases is 0.009 atm
Explanation:
Step 1: Data given
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.
The atmospheric pressure = 0.90 atm
Step 2: Calculate mol fraction
If wehave 100 moles of air, 78 moles will be nitrogen,
21 moles will be oxygen, and 1 mol will be other gases.
Mol fraction = 1/100 = 0.01
Step 3: Calculate the partial pressure of the other gases
Pgas = Xgas * Ptotal
⇒ Pgas = the partial pressure = ?
⇒ Xgas = the mol fraction of the gas = 0.01
⇒Ptotal = the total pressure of the pressure = 0.90 atm
Pgas = 0.01 * 0.90 atm
Pgas = 0.009 atm
The partial pressure of the other gases is 0.009 atm
Answer: The answer is C. The reaction system absorbs 22 kJ of energy from the surroundings.
Explanation: I just took the quiz and it said I got it right.
Either C or D. Those r the answers that make more sense.
