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3 years ago

How to purify ammonia

1 answer:

5 0

In order to purify ammonia to high purity, two basic methods are used. The older one consists in passing gaseous ammonia containing 80 ppm of impurities under atmospheric pressure through liquid ammonia with dissolved metallic sodium.

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For the complete combustion of 47 g of gasoline (octane, C8H18) , the mass of oxygen consumed is

aleksklad [387]

Answer: d) 164.9 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$ $\text{Moles of octane}=\frac{47g}{114g/mol}=0.412moles$

The balanced chemical reaction is:

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)$

According to stoichiometry :

2 moles of $C_8H_{18}$ require = 25 moles of $O_2$

Thus 0.412 moles of $C_8H_{18}$ will require= $\frac{25}{2}\times 0.412=5.15moles$ of $O_2$

Mass of $O_2=moles\times {\text {Molar mass}}=5.15moles\times 32g/mol=164.9g$

Thus 164.9 g of oxygen is consumed.

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3 years ago

If your weight on earth is 107 pound and the force of gravity on the sun is 27 times the gravity of the earth what would your we

Wewaii [24]

Hi! It's simple you just have to multiply 107 times 27 which is 2,889 pounds

5 0

3 years ago

If a molecule contains polar bonds the molecule __ must be polar overall

weeeeeb [17]

Answer:

diatomic

Explanation:

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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:

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

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

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

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

4 0

3 years ago

Write balanced equations and Kb expressions for these Brønsted-Lowry bases in water:Benzoate ion, C₆H₅COO⁻

Dima020 [189]

The balanced equation is,

OBz-(aq) + H2O(l) ---> HOBz(aq) + OH-(aq)

The equation for Kb is :

Kb = [HOBz][OH-]/[OBz]

<h3>What do we know about benzoate ion?</h3>

The conjugate base of benzoic acid, benzoate is the simplest member of the family of benzoates. It consists of a benzoic acid core with a proton missing to give it a charge of -1. It serves as a xenobiotic metabolite for humans as well as a metabolite for plants. It is a benzoic acid's conjugate base.

To learn more about pKb:

brainly.com/question/12193534

#SPJ4

5 0

2 years ago

How to purify ammonia​

How to purify ammonia