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RSB [31]
2 years ago
13

Choose one or more:

Chemistry
1 answer:
wlad13 [49]2 years ago
6 0

Answer:Electrons exhibit interference or diffraction due to scattering objects or apertures around the same size as their de Broglie wavelength. Electrons exhibit a discrete charge. Electrons exhibit a discrete mass. Electrons may tunnel through barriers of nanometer or below thickness

Explanation:

Electrons behave both as matter an waves in accordance with De Broglie principle of wave particle duality hence they possess an associated wavelength and as such undergo diffraction and interference.

Electrons have a known magnitude of mass and charge.

Many studies have also shown that electrons can tunnel through barriers of 1nm thickness or less.

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500 mL of a solution contains 1000 mg of CaCl2. Molecular weight of CaCl2 is 110 g/mol. Specific gravity of the solution is 0. C
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a) 0,2% w/v

b) r=500

c) 0,0182 M

d) 0,0145 m

e) 0,0137 equivalents

Explanation:

a) % w/v means mass of solute in grams per 100 mililiter of solution. Thus:

%w/v= \frac{1,000 g CaCl2}{500mL}×100 = 0,2%w/v

b) Ratio strength is a way to express concentration.  For w/v is in 1g of solute <em>r</em> mililiters of solution have. Thus, r = 500 because we have in the first 1 g of CaCl₂ in 500 mL of solution.

c) Molarity is moles of solute per liter of solution, thus:

1,000 g of CaCl₂ × \frac{1mol}{110g} = 9,09×10⁻³ moles of CaCl₂

500 mL of solution  × \frac{1L}{1000mL} = 0,500 L of solution

M = \frac{9,09x10^{-3} moles }{0,500 L} = 0,0182 M

d) Molality is moles of solute per kg of solution.

Specific gravity is the ratio between density of the solution and density of a reference substance (Usually water). With a specific gravity of 0,8:

kg of solution = 0,500 L of solution × \frac{0,8 kg}{1L} =<em> </em><em>0,625 kg of solution</em>

m = \frac{9,09x10^{-3}moles }{0,625 kg} = 0,0145 m

e)  In a salt, equivalents are the number of moles ables to replace one mole of charge. In CaCl₂ is ¹/₂ because with  ¹/₂ moles of CaCl₂ it is possible to replace 1 mole of charges. Thus, in 1,5 L there are:

1,5 L ×\frac{0,0182 CaCl2 moles}{1L} × \frac{1equivalent}{2 moles} = 0,0137 equivalents

I hope it helps!

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