Answer: I'm going to guess and say it is h2 or o2 I am not sure.
Explanation:
Answer:
The answer to this is
The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters
Explanation:
To solve this we first list out the variables thus
Density of the water = 1.00 g/mL =1000 kg/m³
density of mercury = 13.6 g/mL = 13600 kg/m³
Standard atmospheric pressure = 760 mmHg or 101.325 kilopascals
Therefore from the equation for denstity we have
Density = mass/volume
Pressure = Force/Area and for a column of water, pressure = Density × gravity×height
Therefore where standard atmospheric pressure = 760 mmHg we have for Standard tmospheric pressure= 13600 kg/m³ × 9.81 m/s² × 0.76 m = 101396.16 Pa
This value of pressure should be supported by the column of water as follows
Pressure = 101396.16 Pa = kg/m³×9.81 m/s² ×h
∴ = 10.336 meters
The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters
Answer: use the ideal gas equation, PV = nRT
Explanation:
The ideal gas equation permits you to determine one variable, P, V, n or T, when you know the other two variables.
In this case, your are told you know P, V and T, so you can solve for n:
n = PV / (RT)
Remember this:
R is the universal constant of gases, and you must use the adequate units.
T is the temperature in absolute scales, i.e Kelvin.
Hydrogen H weight: 81
Non-metal
Hydrogen is the simplest element; an atom consists of only one proton and one electron. It is also the most plentiful element in the universe. Despite its simplicity and abundance, hydrogen doesn't occur naturally as a gas on the Earth--it is always combined with other elements.
period 1 group 1
Hydrogen is easily the most abundant element in the universe. It is found in the sun and most of the stars, and the planet Jupiter is composed mostly of hydrogen. On Earth, hydrogen is found in the greatest quantities as water.
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 %