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3 years ago

What is the importance of a chemical symbol?

Chemistry

2 answers:

Snowcat [4.5K]3 years ago

6 0

Answer:

Explanation:

A chemical symbol is a notation of one or two letters representing a chemical element. Its used to identify elements and atoms in a chemical formula easily. Chemical symbols consist of one or two letters, most often derived from the name of the element.

scoray [572]3 years ago

4 0

Answer:

Chemical symbols are used to standardize the 'language of chemistry' and to identify elements and atoms in a chemical formula easily. Chemical symbols consist of one or two letters, most often derived from the name of the element.

Explanation:

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Using the ideal gas law, PV=nRT, where R=0.0821 L atm/mol K, calculate the volume in liters of oxygen produced by the catalytic

AfilCa [17]

Answer:

V = 2.32 Liters

Explanation:

PV = nRT => V = nRT/P

n = 25.8g/122g/mole = 0.21 mole

R = 0.08206 L·atm/mol·K

T = 25.44°C + 273 = 298.44K

P = 2.22 atm (given in problem)

V = (0.21mol)(0.08206 L·atm/mol·K)(298.44K)/(2.22atm) = 2.32 Liters at 25.44°C & 2.22atm

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3 years ago

All of the following statements about different elements are true EXCEPT: Group of answer choices Krypton is one of the noble ga

seropon [69]

Sulfur is an element of the periodic table that is not considered a metalloid.

<h3>What is sulfur?</h3>

Sulfur is a chemical element of the periodic table that has the following characteristics:

Atomic number 16
S symbol
Sulfur is classified as a nonmetal
It has a yellow color

<h3>What are metalloids?</h3>

Metalloids are a set of chemical elements of the periodic table that are characterized by having an intermediate behavior between metals and non-metals, in terms of ionization energies and binding properties.

It is not easy to distinguish them from true metals. They conduct electrical current better than non-metals, but they are not good conductors like metals. In addition, they are usually very varied in their shape and coloration.

The list of metalloids includes the following elements:

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Germanium (Ge)
Arsenic (As)
Antimony (Sb)
Tellurium (Te)
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Learn more about periodic table in: brainly.com/question/11155928

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2 years ago

Cassady? you there ?

iren [92.7K]

Answer:

who da hell is cassady?

Explanation:

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3 years ago

Read 2 more answers

5. I have a mixture of salt, water,

dolphi86 [110]

Iron is left in the filter and salt solution (salt and water) passes into the cup.

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3 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

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3 years ago