Answer:
benefits: can help reduce waste in land fills, and promotes reusability (turned into different products)
drawbacks: can be much more expensive to salvage than to just throw it in a landfill, and most of the time products cannot be recycled effectively due to its material.
What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100
Answer:
A lot of them.
Explanation:
It would take hundreds of thousands of trees to clear all of the emmisions.
Answer:
Rb+
Explanation:
Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.
Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.
Thus our calculations are:
V = 17.0 mL x 1 L / 1000 mL = 0.017 L
2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol
0.0425 mol = 4.36 g/ MW XOH
MW of XOH = (atomic weight of X + 16 + 1)
so solving the above equation we get:
0.0425 = 4.36 / (X + 17 )
0.7225 +0.0425X = 4.36
0.0425X = 4.36 -0.7225 = 3.6375
X = 3.6375/0.0425 = 85.59
The unknown alkali is Rb which has an atomic weight of 85.47 g/mol
Answer:
136.63 °C
Explanation:
ΔTb=Tb solution - Tb pure
Where; Tb pure = 133.60°C
molar mass of solute = 121.14 g/mol
number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles
molality = 0.431 moles/350 * 10^-3 = 1.23 molal
Then;
ΔTb = Kb * m * i
Kb = 2.46°C kg mol^-1
m = 1.23 molal
i = 1
ΔTb = 2.46 * 1.23 * 1
ΔTb = 3.03 °C
Hence;
Tb solution = ΔTb + Tb pure
Tb solution = 3.03 °C + 133.60°C
Tb solution = 136.63 °C
