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boyakko [2]
3 years ago
12

An unknown substance has a melting point of 1064 ∘C, is insoluble in water, conducts electricity as a solid, and is hard. Given

these properties, which of the following are possible identities for the unknown substance? Check all that apply.
Check all that apply.
A. SiO2
B. F2
C. Au
D. KBr
Chemistry
1 answer:
denpristay [2]3 years ago
7 0

Answer:

The unknown substance is Au ( gold)

Explanation:

SiO2 has a high melting point, is insoluble in water but is a very poor conductor. So the statements do not apply here.

F2 is soluble in water, has a melting point way lower than 1064 °C, and isa gas. F2 doesn't apply here.

Au is insoluble in water, has a high melting point, is a good conductor and is hard. Gold applies are all those statements.

Kbr is good soluble in water, and is a bad conductorbecause  it is just a powder, and the molecules have no mobility, and therefore it cannot conduct electricity. The unknown substance is not Kbr.

The unknown substance is Au ( gold)

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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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Answer:

a) 0,2% w/v

b) r=500

c) 0,0182 M

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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

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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

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