[CO] = 1 mol / 2L = 0.5 M

[
According to the equation:

and by using the ICE table:

CO(g) + H2O(g) ↔ CO2(g) + H2(g)

initial 0.5 0.5 0 0

change -X -X +X +X

Equ (0.5-X) (0.5-X) X X

when Kc = X^2 * (0.5-X)^2

by substitution:

1.845 = X^2 * (0.5-X)^2 by solving for X

∴X = 0.26

∴ [CO2] = X = 0.26

4 0

3 years ago

2Na(s) + 2H,O(l) — 2NaOH(?) + H, (g) 0 what is the missing symbol

Klio2033 [76]

Answer:

the symbol that is missing might be 2.

Explanation:

I am not 100% on this, so correct me if I am wrong.

3 0

2 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$