Mass of water produced : 0.146 g
<h3>Further explanation</h3>
Given
33.2 mL of 0.245 M lithium hydroxide
Required
mass of water
Solution
Reaction
HNO₃ (aq) + LiOH (aq) → H₂O (l) + LiNO₃ (aq)
mol LiOH :
= M x V
= 0.245 x 33.2 ml
= 8.134 mmol
From the equation, the mol ratio of HNO₃ : H₂O = 1 : 1, so mol H₂O = 8.134 mmol
mass H₂O :
= mol x MW
= 8.134 x 10⁻³ mol x 18 g/mol
= 0.146 g
Hello!
Numbers from -998 and 999 are in between -999 and 1000.
I hope this was helpful! c:
Answer:
80mL
Explanation:
Step 1:
Data obtained from the question.
Initial Volume (V1) = 40mL
Initial temperature (T1) = –123°C
Final temperature (T2) = 27°C
Final volume (V2) =..?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
T(K) = T(°C) + 273
Initial temperature (T1) = –123°C =
–123°C + 273 = 150K
Final temperature (T2) = 27°C = 27°C + 273 = 300K
Step 3:
Determination of the final volume.
This can be obtained as follow:
V1/T1 = V2/T2
Initial Volume (V1) = 40mL
Initial temperature (T1) = 150K
Final temperature (T2) = 300
Final volume (V2) =..?
V1/T1 = V2 /T2
40/150 = V2 /300
Cross multiply
150 x V2 = 40 x 300
Divide both side by 150
V2 = (40 x 300) /150
V2 = 80mL
Therefore, the new volume of the gas is 80mL
Boyle's Law states: pV = constant.
24.43 x 1.895 = 46.29485
therefore, 15.6 x _____ = 46.29485
unknown = 2.968L