Answer:A.salinity levels, air quality, water density
Explanation:nba youngboy
Explanation:
a process for separating minerals from gangue by taking advantage of the differences in their hydrophobicity.
I believe that is accurate
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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 %
A salt and water I believe
In the given condition where the thylakoids are isolated from the chloroplasts to make the pH of the thylakoid space acidic, the isolated thylakoid will form the ATP.
ATP is also referred to as Adenosine Triphosphate and is correctly called as the energy currency of the cell as it is responsible molecule for carrying the energy units in the cell. When the isolated thylakoids are made to transfer in the pH = 8 basic solution, then the acidic lumen of the chloroplast will be having a higher concentration of H⁺ ions. When they are placed in the basic solution then the H⁺ ions will move out because of the occurrence of concentration gradient which is higher in the lumen than the surrounding. The chloroplasts will now bear synthesis of ATP. Therefore, the isolated thylakoid is able to make ATP when the lumen is acidic, and the surroundings are made basic in nature.
Thus, the thylakoid which is isolated will bear the synthesis of ATP.
Learn more about ATP at:
brainly.com/question/252380
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