Answer:
4000 L
Explanation:
Step 1:
Data obtained from the question. This include the following:
Initial volume (V1) = 2000 L.
Initial temperature (T1) = 100 K.
Initial pressure (P1) = 100 kPa.
Final temperature (T2) = 400 K.
Final pressure (P2) = 200 kPa.
Final volume (V2) =..?
Step 2:
Determination of the new volume of the gas.
The new volume of the gas can be obtained by using the general gas equation as follow:
P1V1/T1 = P2V2/T2
100 x 2000/100 = 200 x V2/400
Cross multiply to express in linear form.
100 x 200 x V2 = 100 x 2000 x 400
Divide both side by 100 x 200
V2 = (100 x 2000 x 400)/(100 x 200)
V2 = 4000 L
Therefore, the new volume of the gas is 4000 L
Answer:
They want to escape from the fury of a hurricane.
Explanation:
Your welcomeeeeeeeeeeeee
The primary mechanism for converting atmospheric carbon into usable forms for life on earth is achieved by photonsynthesis. It is a process where plants convert carbon dioxide into forms that can be used by living organisms which is oxygen. Oxygen is a very essential product of photosynthesis which is being used by living organisms to survive.
Answer:
Mass percent concentration: 25,2% m/m
Explanation:
The percentage mass by mass indicates the grams of solute in 100 grams of solution. We convert the ml of water into g from the density formula:
δ= m/v m=δx v= 1 g/ml x 115 ml= 115 g
115g solution-----29 g NaCl
100g solution----x=(100g solutionx29 g NaCl)/115g solution
x=25,2173913 g NaCl
Q1)
the reaction that takes place is
lead nitrate reacting with potassium iodide to form lead iodide and potassium nitrate
balanced chemical equation for the reaction is as follows
Pb(NO₃)₂ + 2KI ----> PbI₂ + 2KNO₃
Q2)
mass of lead nitrate present - 0.600 g
number of moles = mass present / molar mass
number of moles - 0.600 g / 331.2 g/mol = 0.00181 mol
Q3)
mass of potassium iodide present - 0.850 g
number of moles = mass present / molar mass
number of moles of potassium iodide = 0.850 g / 166 g/mol = 0.00512 mol
Q4)
we have to calculate the number of moles of PbI₂ formed based on the number of moles of Pb(NO₃)₂ present assuming the whole amount of Pb(NO₃)₂ was used up
stoichiometry of Pb(NO₃)₂ to PbI₂ is 1:1
number of Pb(NO₃)₂ moles reacted - 0.00181 mol
therefore number of PbI₂ moles formed - 0.00181 mol
Q5)
next we have to calculate the number of moles of PbI₂ formed based on the amount of KI moles present , assuming all the moles of KI were used up in the reaction
stoichiometry of KI to PbI₂ is 2:1
number of moles of KI reacted - 0.00512 mol
then number of moles of PbI₂ formed - 0.00512 x 2 = 0.0102 mol
0.0102 mol of PbI₂ is formed
Q6)
limting reactant is the reactant that is fully consumed during the reaction. the amount of product formed depends on the amount of limiting reactant present
if lead nitrate is the limiting reactant
if 1 mol of Pb(NO₃)₂ reacts with 2 mol of KI
then 0.00181 mol of Pb(NO₃)₂ reacts with - 2 x 0.00181 mol of KI = 0.00362 mol
but 0.00512 mol of KI is present and only 0.00362 mol are required
therefore KI is in excess and Pb(NO₃)₂ is the limiting reactant
Pb(NO₃)₂ is the limiting reactant
Q7)
then the amount of PbI₂ formed depends on amount of Pb(NO₃)₂ present
therefore number of moles of PbI₂ formed is based on number of Pb(NO₃)₂ moles present
as calculated in Question number 4 - Q4
number of PbI₂ moles formed - 0.00181 mol
mass of PbI₂ formed - 461 g/mol x 0.00181 mol = 0.834 g
mass of PbI₂ formed - 0.834 g
Q8)
actual yield obtained is not always equal to the theoretical yield . therefore we have to find the percent yield. This tells us the percentage of the theoretical yield that is actually obtained after the experiment
percent yield = actual yield / theoretical yield x 100 %
percent yield = 0.475 g / 0.834 g x 100 % = 57.0 %
percent yield of lead iodide is 57.0 %
