Answer:
Hazardous types: This type poses potential threats to the environment and human life.Battery wastes from thrown away technology.
Electronic waste or e-waste describe discarded electrical devices. Used electronics which are destined for refurbishment, resale, salvage recycling through material recovery, or disposal are also considered e-waste.
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Answer:
338.00 mL
Explanation:
The lead ions come from the salt Pb(NO₃)₂ and the iodide from the acid HI, so the balanced reaction is:
Pb(NO₃)₂(aq) + 2HI(aq) → PbI₂(s) + 2HNO₃(aq)
So, the stoichiometry is 1 mol of Pb(NO₃)₂ to 2 moles of HI, then:
1 mol of Pb(NO₃)------------------------------------ 2 moles of HI
0.600 mol of Pb(NO₃)₂--------------------------- x
By a simple direct three rule:
x = 1.200 mol of HI
The acid has concentration equal to 3.550 mol/L, the volume (V) is the number of moles divided by the molar concentration:
V = 1.200/3.550 = 0.338 L
V = 338.00 mL
Answer:
matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants. Balanced chemical equations show that mass is conserved in chemical reactions.
Explanation:
Answer:
The correct answer is - 800.
Explanation:
Given:
Total amount = ? or assume x
spend in buying birthday item = 3/4 of x
given to sister = 1/5 of x
remaining to mother = 40
solution:
the remaning amount = x- (3x/4+x/5) = 4=
=> x- 19x/20 = 40
=> x = 20*40
=> x = 800
thus, the correct answer is = 800
Answer:
87.9%
Explanation:
Balanced Chemical Equation:
HCl + NaOH = NaCl + H2O
We are Given:
Mass of H2O = 9.17 g
Mass of HCl = 21.1 g
Mass of NaOH = 43.6 g
First, calculate the moles of both HCl and NaOH:
Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles
Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles
Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:
Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles
Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles
From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:
Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g
% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%