I don't know but I would pick b.2.covalent bonds
Answer:
4.37 g of barium sulphate
Explanation:
The reaction equation is;
3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)
From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles
To find the limiting reactant;
3 moles of barium chloride yields 3 moles of barium sulphate
0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate
1 mole of iron III sulphate yields 3 moles of barium sulphate
0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate
Hence,barium chloride is the limiting reactant
Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate
Answer:
6 x 10⁵ kg Hg
Explanation:
The mass of mercury in the entire lake is found by multiplying the concentration of the mercury by the volume of the lake.
The volume of the lake is calculated in cubic feet:
V = (SA)x(depth) = (100mi²)(5280ft/mi)² x (20ft) = 5.57568 x 10¹⁰ ft³
Cubic feet are then converted to mL (1cm³=1mL)
(5.57568 x 10¹⁰ ft³) x (12in/ft)³ x (2.54cm/in)³ = 1.578856752 x 10¹⁵ mL
The mass of mercury is then found:
m = CV = (0.4μg/mL)(1g/10⁶μg)(1kg/1000g) x (1.578856752 x 10¹⁵ mL) = 6 x 10⁵ kg Hg
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