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3 years ago

What does it mean that something is a conservative ion?

Chemistry

1 answer:

mario62 [17]3 years ago

5 0

Answer:

The major ionic constituents whose concentrations can be determined from the salinity are known as conservative substances. Their constant relative concentrations are due to the large amounts of these species in the oceans in comparison t

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Nickel has an FCC structure and an atomic radius of 0.124 nm. (12 points) a. What is the coordination number? b. Calculate the A

LekaFEV [45]

Answer:

a)CN=12

b)APF=74 %

c)a=0.35 nm

d)ρ=9090.9 $Kg/m^3$

Explanation:

Given that

Nickel have FCC structure

We know that in FCC structure ,in FCC 8 atoms at corner with 1/8 th part in one unit cell and 6 atoms at faces with 1/2 part in one unit cell .

Z=8 x 1/8 + 1/2 x 6 =4

Z=4

Coordination number (CN)

The number of atoms which touch the second atoms is known as coordination number.In other word the number of nearest atoms.

CN=12

Coordination number of FCC structure is 12.These 12 atoms are 4 atoms at the at corner ,4 atoms at 4 faces and 4 atoms of next unit cell.

APF

$APF=\dfrac{Z\times \dfrac{4}{3}\pi R^3}{a^3}$

We know that for FCC

$4R=\sqrt{2}\ a$

Now by putting the values

$APF=\dfrac{4\times \dfrac{4}{3}\pi R^3}{\left(\dfrac{4R}{\sqrt2}\right)^3}$

APF=0.74

APF=74 %

$4R=\sqrt{2}\ a$

$4\times 0.124=\sqrt{2}\ a$

a=0.35 nm

Density

$\rho=\dfrac{Z\times M}{N_A\times a^3}$

We know that M for Ni

M=58.69 g/mol

a=0.35 nm

$N_A=6.023\times 10^{23}\ atom/mol$

$\rho=\dfrac{Z\times M}{N_A\times a^3}$

$\rho=\dfrac{4\times 58.69}{6.023\times 10^{23}\times( 0.35\times 10^{-9})^3}\ g/m^3$

ρ=9090.9 $Kg/m^3$

3 0

3 years ago

Which element would most likely have a positive electron affinity?

krok68 [10]

I believe the correct answer from the choices listed above is the first option. The element that would most likely have a positive electron affinity would be sodium, Na. Electron affinity is defined as the change in energy of a neutral atom when an electron is added to the atom.

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3 years ago

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Write a letter to your friend describing your feelings after you left the school

fredd [130]

Dear friend,
How are you doing? I finally switched to my new school and it's been so hard adjusting. But I know I have a new chance at making new friends and memories. Also, I can learn new things, like interacting with new people and how to keep my anxiety low. I really miss us hanging out and how much we'd laugh and get in trouble at class. But i really hope to see you soon & that it hasn't gone terribly wrong for you since i left.
your friend,
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3 years ago

Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

Luden [163]

Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

Molar mass of $NaN_3$ = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

$n=0.7015moles$

The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

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3 years ago

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Angelina_Jolie [31]

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3 years ago

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