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2 years ago

A compound contains 34.5% calcium, 24.1% silicon and 41.4% oxygen by mass. What is its empirical formula?

Chemistry

1 answer:

Amiraneli [1.4K]2 years ago

4 0

<em> $empirical \: formula \\ = CaSiO3 \\ please \: see \: the \: attached \: picture \\ hope \: it \: helps$ </em>

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Which phrase is the best definition of matter?

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The amount of physical material in an object

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2 years ago

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Write the electron configuration (ground and excited states) for element with atomic number 16 ?

Talja [164]

2, 8,6 because it has to be in a configuration of 2,8,8

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3 years ago

The substances sodium (Na), oxygen (O2), and sodium oxide (Na20) participate in this chemical reaction

Margaret [11]

Answer:

Explanation:

This is a synthesis or a combination reaction , which involves the oxygen combining with sodium to make one compound.

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3 years ago

Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC

sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

$%IC = (1 - e^{-0.25*(x_A - x_B)^2})$ *100%

Where $x_A - x_B$ is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. $x_{Ti}$ = 1.5

$x_{O}$ = 3.5

$%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})$ *100%

%IC = 63.2%

b. $x_{Zn}$ = 1.6

$x_{Te}$ = 2.1

$%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})$ *100%

%IC = 11.7%

c. $x_{Cs}$ = 0.7

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})$ *100%

%IC = 73.3%

d. $x_{In}$ = 1.7

$x_{Sb}$ = 1.9

$%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})$ *100%

%IC = 0.995 %

e. $x_{Mg}$ = 1.2

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})$ *100%

%IC = 55.5%

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3 years ago

Describe the evidence of changes in organisms in the Cambrian Era that are reflected in the fossil record.

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Answer:

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Explanation:

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2 years ago