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2 years ago

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Lithium ions in Lithium selenide (Li2Se) have an atomic radius of 73 pm whereas the selenium ion is 184 pm. This compound is mos

t likely to adopt a:Select the correct answer below:
a. closest-packed array with lithium ions occupying tetrahedral holes
b. closest-packed array with lithium ions occupying octahedral holes
c. body-centered cubic array with lithium ions occupying cubic holes
d. none of the above

1 answer:

NARA [144]2 years ago

3 0

Explanation:

Formula according to the radius ratio rule is as follows.

$\frac{r_{+}}{r_{-}} = \frac{73}{184}$

= 0.397

According to the radius ratio rule, as the calculated value is 0.397 and it lies in between 0.225 to 0.414. Therefore, it means that the type of void is tetrahedral.

Thus, we can conclude that the given compound is most likely to adopt closest-packed array with lithium ions occupying tetrahedral holes.

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Answer:

$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

Explanation:

Energy of the photon can be calculated by

$E=h\nu=\frac{h\times c}{\lambda}$ (Planck's equation)

where,

E = energy of photon

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$E=\frac{(6.63\times 10^{-34}Js)\times (3\times 10^8m/s)}{510\times 10^{-9}m}$

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$4\times 10^{-9} J$ is the approximate energy of one photon of this light.

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2 years ago

HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.

Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, $C_A$ = 6.0mol/dm³

volume of HA, $V_A$ = 25.0cm³, = 0.025dm³

Concentration of HB, $C_B$ = 3.0mol/dm³

volume of HB, $V_B$ = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

$C_iVi = C_fV_f$

where;

$C_i$ is initial concentration

$V_i$ is initial volume

$C_f$ is final concentration of the solution

$V_f$ is final volume of the solution

$C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3$

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Answer

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