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I mole of sulfur atoms is 32 g. Thus there are 300/32 moles in 300 g, i.e. 9.375 moles.
50ml = .05L of HCL 1.0 M = 1mol / 1L of HCL simply multiply - .05 by 1.0, and get your answer!
the answer is 1.5
At low concentrations, normality and molarity are about the same. So figure out how many grams are in a mole of KOH: K-39, O-16, H-1, so 39+16+1=56g/mol. 0.05 moles would be 56(.05)=2.8g, so dissolve 2.8 grams of KOH in a liter of water and you're there.
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Moles of the soild = 1.78/ 208 = .00856 moles of H2O = 0.90/ 18 =.05 therefore number of moles of water for 1 mole of solid = .05/.00856 = approx 6
I mole of sulfur atoms is 32 g. Thus there are 300/32 moles in 300 g, i.e. 9.375 moles.
137.4 g First you need to calculate the molar mass of the molecule. Be = 9.01 g/mol C = 12.01 g/mol I = 126.9 g/mol Thus BeCI2 is equivalent to ( 9.01 + 12.01 + 2(126.9) ) g/mol or 274.82 g/mol. Then, using conversions, you multiply the amount you have by the molar mass so that: (.05 mol BeCI2)*(274.82 g/mol). The moles cancel out and you are left with a weight of 137.41 g.
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M = g/L 20 = g/0.05 20 * .05 = g 1.0 = g of solute
50ml = .05L of HCL 1.0 M = 1mol / 1L of HCL simply multiply - .05 by 1.0, and get your answer!
4.5 * 10^15 J
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