The chemical reaction would be:
HBr + LiOH = LiBr + H2O
We are given the concentration of the other reactant. We use these given values and the reaction to relate to moles of HBr needed. We do as follows:
0.253 mol / L ( .005 L) ( 1 mol HBr / 1 mol LiOH ) = 0.0013 mol HBr
Molarity = 0.0013 / 0.01 = 0.13 M
Answer:
27.9 g
Explanation:
CsF + XeF₆ → CsXeF₇
First we <u>convert 73.1 g of cesium xenon heptafluoride (CsXeF₇) into moles</u>, using its<em> molar mass</em>:
- Molar mass of CsXeF₇ = 397.193 g/mol
- 73.1 g CsXeF₇ ÷ 397.193 g/mol = 0.184 mol CsXeF₇
As <em>1 mol of cesium fluoride (CsF) produces 1 mol of CsXeF₇</em>, in order to produce 0.184 moles of CsXeF₇ we would need 0.184 moles of CsF.
Now we <u>convert 0.184 moles of CsF to moles</u>, using the <em>molar mass of CsF</em>:
- Molar mass of CsF = 151.9 g/mol
- 0.184 mol * 151.9 g/mol = 27.9 g
Energy is used by plants to carry cellular respiration.
Answer:
2
Explanation:
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