The chemical reaction would be expressed as follows:
HBr + LiOH = LiBr + H2O
We are given the volumes and corresponding concentration to be used for the reaction. We use these values to solve for the concentration of the other reactant. We do as follows:
0.253 mol LiOH / L solution ( 0.01673 L ) ( 1 mol HBr / 1 mol LiOH ) = 0.00423 HBr needed
Concentration of HBr =0.00423mol / .010 L = 0.423 M HBr
Idk i just need to ask a question
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Answer:
The answers to your question are below
Explanation:
a) 6.85×1020 H2O2 molecules
H2O2 MW = 32 + 2 = 34 g
34g -------------------- 6.023 x 10²³ molecules
x ------------------- 6.85 x 10 ²⁰
x = (6.85 x 10 ²⁰)(34)/ 6.023 x 10²³
x = 0.038 g
3.3×1022 SO2 molecules
MW SO2 = 32 + 32 = 64g
64 g -------------------- 6.023 x 10²³ molecules
x -------------------- 3.3×1022 SO2 molecules
x = (3.3×1022 SO2)(64) / 6.023 x 10²³
x = 3.51 g
5.5×1025 O3 molecules
MW = 16 x 3 = 48g
48 g ----------------- 6.023 x 10²³ molecules
x ------------------ 5.5×1025 O3 molecules
x = (5.5×1025 )(48) / 6.023 x 10²³
x = 4383 g
9.30×1019 CH4 molecules
MW = 12 + 4 = 16 g
16 g -------------------- 6.023 x 10²³ molecules
x -------------------- 9.30×1019 CH4 molecules
x = (9.30×1019)(16) / 6.023 x 10²³
x = 0.0025 g
