Answer:
625.46 °C
Explanation:
We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 19 °C
T(K) = 19 °C + 273
T(K) = 292 K
Next, we shall determine the Final temperature. This can be obtained as follow:
Initial volume (V₁) = 3.25 L
Initial temperature (T₁) = 292 K
Final volume (V₂) = 10 L
Final temperature (T₂) =?
V₁/T₁ = V₂/T₂
3.25 / 292 = 10 / T₂
Cross multiply
3.25 × T₂ = 292 × 10
3.25 × T₂ = 2920
Divide both side by 3.25
T₂ = 2920 / 3.25
T₂ = 898.46 K
Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 898.46 K
T(°C) = 898.46 – 273
T(°C) = 625.46 °C
Therefore the final temperature of the gas is 625.46 °C
The arrows represent the movement of starting substances
6.02214086 x 10^23 mol^-1
Answer:
a) Se²⁻> S²⁻ > O²
b) Te²⁻ > I- >Cs+
c) Cs+ > Ba²⁺ > Sr²⁺
Explanation:
(a) Se²⁻, S²⁻, O²⁻
In general, ionic radius decreases with increasing positive charge.
As the charge on the ion becomes more positive, there are fewer electrons.
The ion has a smaller radius. In general, ionic radius increases with increasing negative charge.
For ions of the same charge (e.g. in the same group) the size increases as we go down a group in the periodic table
Se²⁻> S²⁻ > O²
(b) Te²⁻, Cs⁺, I⁻
Te²⁻ > I- >Cs+
Te2- hast the biggest size, because of the double negative charge.
Cs+ has the smallest size since it has the most positive charge, compared to Te2- and I-.
(c) Sr²⁺, Ba²⁺, Cs⁺
Cs+ > Ba²⁺ > Sr²⁺
Cs+ has the biggest size, because its more downward (compared to Sr2+) and more to the left (compared) ot Ba2+.
Sr2+ has the smallest size because it's more upwords (compared to Cs+ and Ba2+)
Answer:
Explanation:
To find the concentration; let's first compute the average density and the average atomic weight.
For the average density 
; we have:
The average atomic weight is:
So; in terms of vanadium, the Concentration of iron is:
From a unit cell volume 
where;
= number of Avogadro constant.
SO; replacing with 
; with 
; 
with 
and
with 
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) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2FA_%7BFe%7D%20%5D%20%2B%20%5BC_v%2FA_v%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2F%5Crho_%7BFe%7D%20%5D%20%2B%20%5BC_v%2F%5Crho_v%5D%7D%20%5CBig%29%20%20%7D)
![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) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100-C_v%29A_%7Bv%7D%20%5D%20%2B%20%5BC_v%2FA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100-C_v%29%2F%5Crho_%7Bv%7D%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
![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) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100A_%7Bv%7D-C_vA_%7Bv%7D%29%20%5D%20%2B%20%5BC_vA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100%5Crho_%7Bv%7D%20-%20C_v%20%5Crho_%7Bv%7D%29%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
Replacing the values; we have:
