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3 years ago

How is the periodic table of the elements arranged?

Chemistry

2 answers:

Bogdan [553]3 years ago

6 0

Answer:

Answer is B

Explanation:

i just took the test

vagabundo [1.1K]3 years ago

3 0

Answer:

B. By aligning groups with common properties.

Explanation:

The periodic table is a table of elements arranged into rows called periods and columns called groups.
Elements with similar chemical properties are placed in the same group. For example, group 1 elements known as alkali metals share similar properties.
Elements in the same period or low have the same number of energy shells or energy levels. For example elements in period 3 have 3 energy levels.

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How does a cell obtain energy

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A process called respiration. This process is carried out in organelles in the cell<span> called the Mitochondria.</span>

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if you find that a substance turns litmus paper to blue and contains hydroxide ions, it is most likely

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4 years ago

Read 2 more answers

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:

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

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

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

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

4 0

3 years ago

If I react 6 units of AB with 10 units of CD in the equation below, what is the limiting reactant? 3AB + 4CD --> 2AD + 6CB

irga5000 [103]

If 6 units of AB were reacted with 10 units of CD in the described equation, the limiting reactant would AB

<h3>Limiting reactants</h3>

They are reactants that determine how far a reaction can go in terms of yield.

From the equation: 3AB + 4CD --> 2AD + 6CB

The mole ratio of AB to CD is 3:4

Thus, 6 units of AB will require 8 units of CD.

But 10 units of CD were reacted with only 6 units of AB. This means that CD is in excess by 2 units while AB will limit the yield of the reaction.

More on limiting reactants can be found here: brainly.com/question/14225536

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2 years ago

How many atoms of Mg are present in 97.22 grams of Mg?

Hitman42 [59]

Answer:

b.......

Explanation:

b.....

5 0

4 years ago