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3 years ago

Add me on insta? Free points btw

Chemistry

2 answers:

pav-90 [236]3 years ago

6 0

Answer:

add uu on insta lemme see about that rq........

lemme call my agent to make sure i can *beep bop bo beep*

okay, what's tha insta "oh ShOot my agent caught gtg bt drop tha insta" .....

hammer [34]3 years ago

4 0

Answer:

i'm down for that

Explanation:

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Stannum has a body centered tetragonal with lattice constant, a = b = 5.83A and c = 3.18A. If the atomic radius is 0.145 nm, det

Fed [463]

Answer:

the atomic packing factor of Sn is 0.24

Explanation:

a = b = 5.83A and c = 3.18A.

Volume of unit cell = a²c

= (5.83)² * 3.18 * 10⁻²⁴ cm³

= 1.08 * 10⁻²²cm³

Volume of atoms =

$2 \times \frac{4}{3} \pi r^3$

(∴ BCC, effective number of atom is 2)

Volume of atoms =

$2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3$

= 2.55*10⁻²³cm³

$\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}$

$=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24$

<h3>therefore, the atomic packing factor of Sn is 0.24</h3>

4 0

3 years ago

What is the formula for volume of a rectangular solid?

WINSTONCH [101]

Answer:

1. length x width x height

5 0

2 years ago

Read 2 more answers

For the following reaction, calculate how many moles of NO2 forms when 0.356 moles of the reactant completely reacts.

Nesterboy [21]

Answer:

0.712 mol of NO₂ are formed .

Explanation:

For the reaction , given in the question ,

2 N₂O₅ ( g ) → 4 NO₂ ( g ) + O₂ ( g )

From the above balanced reaction ,

2 mol of N₂O₅ reacts to give 4 mol of NO₂

Applying unitary method ,

1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂

From the question , 0.356 mol of N₂O₅ are reacted ,

<u>now, using the above equation , to calculate the moles of the NO₂ , as follow -</u>

Since ,

1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂

0.356 mol of N₂O₅ reacts to give 4 / 2 * 0.356 mol of NO₂

Calculating ,

0.712 mol of NO₂ are formed .

7 0

3 years ago

A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.

Reika [66]

Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

HX + H₂O ⇄ X⁻ + H₃O⁺

Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

2H₂O ⇄ H₃O⁺ + OH⁻

Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

Kw / Ka = Kb

1x10⁻¹⁴ / 2.76x10⁻⁸ =

3.62x10⁻⁷ = Kb

4 0

3 years ago

For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80

Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

$A\rightleftharpoons B$

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$

Thus, by recalling the Van't Hoff's equation, we can write:

$ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Hence, we solve for the enthalpy change as follows:

$\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }$

Finally, we plug in the numbers to obtain:

$\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}$

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5 0

3 years ago