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3 years ago

6

Based on the current atomic theory, most chemical reactions take place as a result of the interactions of which sub-atomic parti

cles?

1 answer:

evablogger [386]3 years ago

8 0

Electrons boiiiiiiiiiiiiiiiiiiiiiiiiiii

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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

A solution contains 0.775 g Ca2 0.775 g Ca2 in enough water to make a 1925 mL1925 mL solution. What is the milliequivalents of C

Romashka [77]

Answer: There will be 0.00002 meq per Liter of the solution.

Explanation:

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.

$Normality=\frac{\text{no of gram equivalents} \times 1000}{\text{Volume in ml}}$

$Normality=\frac{\text{given mass}\times 1000}{\text {Equivalent mass}\times {\text{Volume in ml}}}$

Equivalent weight is calculated by dividing the molecular weight by n factor. ${\text{Equivalent weight}}=\frac{\text{Molecular weight}}{n}$

n= charge for charged species , For $Ca^{2+}$ , n =2

${\text{Equivalent weight}}=\frac{40}{2}=20g$

$\text{no of gram equivalents}=\frac{0.775g}{20g/mol}=0.04gramequivalents$

$Normality=\frac{0.04\times 1000}{1925}=0.02eq/L=0.00002meq/L$

Thus there will be 0.00002 meq per Liter of the solution.

3 0

3 years ago

Which of the following is true about a municipal bond with a put option? (A) An investor will exercise the option to put the bon

vampirchik [111]

Answer:

(A) An investor will exercise the option to put the bond if yields rise significantly

Explanation:

A put option on the bond is a mechanism to allow the buyer of the bond the ability to compel the lender to repay the principal on the bond. The put option offers the buyer of the bond the ability to collect the principal of the bond anytime they choose until maturity for any purpose.

Recall that once the price drops (that is, the yield increases), put options are exercised. If the yield significantly increased, the put choice on a municipal bond is executed.

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3 years ago

In the reaction of Mg (s) + 2 HCl (aq) Imported Asset MgCl2 (aq) + H2 (g), what

Pavlova-9 [17]

To decrease the rate of the forward reaction you can do the following:

- decrease the concentration of the reactants

- increase the concentration of the products

- increase the pressure so the hydrogen gas will not be easily removed from the reaction

- decrease the temperature, number of collisions will decrease so the rate of the reaction will decrease

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3 years ago

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[Ar]4s23d104p3 is the electron configuration of a(n) ________ atom.

madreJ [45]

<span>[Ar]4s23d104p3 is the electron configuration of a(n) __As______ atom.</span>

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3 years ago

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