Given, the density of water is 0.9975 g/ml. Density of water is mass of water per unit volume. Mass of 1 ml of water supposed to be 0.9975 g from density of water. So, mass of 10 ml of water is (0.9975 X 10) g= 9.975 g. From graduated cylinder, mass of 10 ml water is measured to be 9.955 g. So, error for mass of 10 ml water= (9.975-9.955)=0.02 g. Percentage of error for 10 ml water is 
= 0.2. Error in the mass for the 10 ml of water is 0.2 %.
Answer:
2C2H6 + 7O2 → 4 CO2 + 6H20
Explanation:
cyanoacrylate
When cyanoacrylate polymerizes, the cyanoacrylate sticks together on an atomic level, and it becomes solid. Super Glue solidifies in all the microscopic crevices of an object that it moved into when it was a liquid, and it can form chemical bonds with any anions in the object.
Answer:
7800 cells
Explanation:
The number of Escherichia coli cells N after n generations is given by
N = 2ⁿ × N₀ where N₀ = initial population
Now making N₀ subject of the formula, we have
N₀ = N/2ⁿ
Since n = 8 and N = 1,996,800 cells, substituting the values of the variables into the equation, we have
N₀ = N/2ⁿ
N₀ = 1,996,800 cells/2⁸
N₀ = 1,996,800 cells/256
N₀ = 7800 cells
So, the initial population of Escherichia coli cells is 7800 cells.
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 %
