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kifflom [539]
4 years ago
12

Describe when an ionic compound can conduct electricity

Chemistry
1 answer:
givi [52]4 years ago
7 0
They cannot move to conduct the electric current. But when an ionic compound melts, the charged ions are free to move. Molten ionic compounds do conduct electricity. When a crystal of an ionic compound dissolves in water, the ions separate.
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Based on the chart below, which of the tests measured physical properties?
sp2606 [1]
Density and phase are both physical properties
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3 years ago
Which class of elements will contain both solids and gases at room temperature?
Ainat [17]
The nonmetals, as s class of elements contains both solids and gases at room temperature. 
Some gases classed as nonmetals are : hydrogen, helium, oxygen, nitrogen, fluorine, neon, or radon and many others. An example of a solid that is a nonmetal is sulfur.
5 0
4 years ago
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A 20.0 mL sample of glycerol has a mass of 25.2 grams. What is the mass of a 57 mL sample of glycerol?
dsp73

Answer:

Density = 25.2 g/20.0 ml = 1.26 g/ml

57 ml x 1.26 g/ml = 71.8 g

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4 years ago
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Which states of matter have particles that move independently of one another with very little attraction?
bonufazy [111]

Answer:

Gas

Explanation:

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4 0
3 years ago
Nicotine, a component of tobacco, is composed of c, h, and n. A 2.625-mg sample of nicotine was combusted, producing 7.1210 mg o
Vikki [24]

The mass sample of nicotine combusted = 2.625 mg  (given)

Mass of CO_2 produced = 7.1210 mg  (given)

Mass of H_2O produced = 2.042 mg  (given)

Molar mass of CO_2 = 44 g/mol

Molar mass of H_2O = 18 g/mol

Percentage of Carbon = \frac{12 g}{44 g/mol}\times \frac{7.120 mg of CO_2}{2.625 mg sample}\times 100 = 74.04%

Percentage of hydrogen = \frac{2 g}{18 g/mol}\times \frac{2.04 mg of H_2O}{2.625 mg sample}\times 100 = 8.62%

Now, for percentage of nitrogen = 100 - (74.04+8.62) = 17.34%

Calculating the moles of each element:

Number of moles = \frac{given mass}{Molar mass}

  • For C

\frac{74.04 g}{12 g/mol} = 6.17 mol

  • For H

\frac{8.62 g}{1 g/mol} = 8.62 mol

  • For N

\frac{17.34 g}{14 g/mol} = 1.24 mol

Dividing with the smallest mole value to calculate the molar ratio of each element:

C_{\frac{6.17}{1.24}} = C_{4.9} = C_5

H_{\frac{8.62}{1.24}} = H_{6.9} = H_7

N_{\frac{1.24}{1.24}} = N_1

Hence, the empirical formula for nicotine is C_5H_7N.

8 0
3 years ago
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