Answer:
48.84mL
Explanation:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the question:
nA = 1
nB = 2
From the question given we obtained the following information:
Ma = 0.43M
Va =?
Mb = 0.35M
Vb = 120mL
Using MaVa / MbVb = nA/nB, we can easily find the volume of the acid required. This is illustrated below:
MaVa / MbVb = nA/nB
0.43 x Va / 0.35 x 120 = 1/2
Cross multiply to express in linear form
2 x 0.43 x Va = 0.35 x 120
Divide both side by the (2 x 0.43)
Va = (0.35 x 120) /(2 x 0.43)
Va = 48.84mL
Therefore, the volume of H2SO4 required is 48.84mL
The most reasonable step to take will be to check the calibration of the
thermometer to ensure that it is giving accurate readings.
Experiments which are done to be give the same results under standard
conditions. Common causes of incorrect results during the analysis of
boiling point of substances include heating the substance too rapidly or
the thermometer not being properly calibrated.
The thermometer is used to measure the temperature and so must be
properly calibrated to give accurate readings.
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A
"The heat from the hot chocolate will travel to the spoon"
Answer:
The answer to your question is V2 = 434.7 l
Explanation:
Data
Volume 1 = V1 = 240 l Volume 2 = ?
Temperature 1 = T1 = 479°K Temperature 2 = T2 = 293°K
Pressure 1 = P1 = 300 KPa Pressure 2 = P2 = 101.325 Kpa
Process
1.- Use the combined gas law to solve this problem
P1V1/T1 = P2V2/t2
-Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (300)(240)(293) / (479)(101.325)
3.- Simplification
V2 = 21096000 / 48534.675
4.- Result
V2 = 434.7 l
Answer:
1) The rate of the overall reaction = Δ[N₂O]/Δt = 0.015 mol/L.s.
2) The rate of change for NO = - Δ[NO]/Δt = 3 Δ[N₂O]/Δt = 0.045 mol/L.s.
Explanation:
<em>3NO(g) → N₂O(g) + NO₂(g).</em>
The rate of the reaction = -1/3 Δ[NO]/Δt = Δ[N₂O]/Δt = Δ[NO₂]/Δt.
Given that: Δ[N₂O]/Δt = 0.015 mol/L.s.
<em>1) The rate of the overall reaction is?</em>
The rate of the overall reaction = Δ[N₂O]/Δt = 0.015 mol/L.s.
<em>2) The rate of change for NO is?</em>
The rate of change for NO = - Δ[NO]/Δt.
∵ -1/3 Δ[NO]/Δt = Δ[N₂O]/Δt.
<em>∴ The rate of change for NO = - Δ[NO]/Δt = 3 Δ[N₂O]/Δt </em>= 3(0.015 mol/L.s) = <em>0.045 mol/L.s.</em>
