All categories

3 years ago

What is the PEN number for Oxygen?

Chemistry

1 answer:

nataly862011 [7]3 years ago

7 0

Answer:

Explanation:

g o o g l e d it

You might be interested in

At liftoff, a space shuttle uses a solid mixture of ammonium perchlorate and aluminum powder to obtain great thrust from the vo

klemol [59]

1 - A balanced equation for the reaction is 3NH4ClO4 + 3Al -------> Al2O3 + AlCl3 + 6H2O + 3NO

2 - Ammonium perchlorate is the oxidizing agent while aluminum powder is the reducing agent.

Molar mass of aluminum oxide = 102 g/mol;

The molar mass of ammonium perchlorate =117.5 g/mol.

From the equation of reaction 3 moles of ammonium perchlorate react with 3 moles of aluminum to produce 1 mole of aluminum oxide.

That is 3 x 7.5 g of ammonium perchlorate reacts with g of aluminum to produce 102 g/mol of aluminum oxide. Also 150g.

150 g of ammonium perchlorate produces (150 * 102)/(3*117.5) g of aluminum oxide = 43.4 g of aluminum oxide.

If you need to learn more about the balanced equation, click here;

brainly.com/question/26694427

#SPJ4

5 0

2 years ago

In this image, which fundamental interaction is responsible for attracting the electron?

Elza [17]

Answer:

Explanation:

electromagnetism

8 0

3 years ago

Read 2 more answers

PLZZZZ HELP ASAP!!! Sharpest pH change

Lynna [10]

A example it a because titration is weak Suck it greeeeeeeeennnnnn boyyyyyyyyyy

6 0

3 years ago

On the Periodic Table, which elements are predominantly (mostly) gases with low densities?

aliya0001 [1]

I’m sorry if I wasted your time but I think it’s alkali metals but I’m
Not sure

5 0

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:

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

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

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

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

4 0

3 years ago