334 g x 1 mol/331.6 g x 6.02x10^23 molecules/mole = answer
325 g of CBr4 have 5,9.10e23 molecules.
To answer this, you need to know the ∆Hfusion of water, which happens to be 334 J/g. So, to melt 12.8 g of ice at 0ºC, the joules needed = (12.8 g)(334 J/g) = 4275 joules
2.05×1022 molecules
The answer is 8,5379.10e23 molecules.
The number of molecules is 6,3985.10e24.
325 g of CBr4 have 5,9.10e23 molecules.
The molecular mass of CBr4 is 12.0 + 4(79.9) = 331.6Amount of CBr4 = mass of substance / molecular mass = 393/331.6 = 1.19mol This means that a 393g pure sample contains 1.19 moles of tetrabromomethane. The Avogadro's number is 6.02 x 10^23 So, number of molecules of CBr4 = 1.19 x 6.02 x 10^23 = 7.13 x 10^23
To answer this, you need to know the ∆Hfusion of water, which happens to be 334 J/g. So, to melt 12.8 g of ice at 0ºC, the joules needed = (12.8 g)(334 J/g) = 4275 joules
22.0 g of silver chloride contain 0,918.10e23 molecules.
26.3 g of calcium hydroxide contain 2,054 molecules.
32 g SO2 x 1 mole SO2/96 g x 6.02x10^23 molecules/mole = 2.0x10^23 molecules
The answer is 8,5379.10e23 molecules.
The number of molecules is 6,3985.10e24.
2.05×1022 molecules
The answer is 1,57.10e27 molecules.
1,4.10e23 molecules
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