Answer:
hola soy jess, tu respuesta esta aqui
¿cuantos moles de CO2 se requiere para reaccionar 2 moles de Ba(OH)2
2 mol Ba(OH)₂ × \frac{1molCO_{2} }{1molBa (OH)_{2}}
1molBa(OH)
2
1molCO
2
= 2 moles CO₂
Explanation:
espero que pueda ayudarte
hermana/hermano
lo que
hahahaha
To solve this problem, we must assume ideal gas behaviour so
that we can use Graham’s law:
vA / vB = sqrt (MW_B / MW_A)
where,
<span>vA = speed of diffusion of A (HBR)</span>
vB = speed of diffusion of B (unknown)
MW_B = molecular weight of B (unkown)
MW_A = molar weight of HBr = 80.91 amu
We know from the given that:
vA / vB = 1 / 1.49
So,
1/1.49 = sqrt (MW_B / 80.91)
MW_B = 36.44 g/mol
Since this unknown is also hydrogen halide, therefore this
must be in the form of HX.
HX = 36.44 g/mol , therefore:
x = 35.44 g/mol
From the Periodic Table, Chlorine (Cl) has a molar mass of
35.44 g/mol. Therefore the hydrogen halide is:
HCl
At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2
at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root:
∴ √(1/7.04X10^-2)= [HBr] / [H2]^1/2*[Br]^1/2
∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769
ANSWERS:
1.
⇒ 2KCl +
2.
⇒
3.
⇒
4.
⇒
5.
⇒
I have attached my work for 1-3. I would like to see if you can get 4 and 5 on your own. but, if you are struggling/copnfused please let me know in the comments! :)
Answer:
The volume increases by 100%.
Explanation:
<u>Step 1:</u> Data given
Number of moles ideal gas = 1 mol
Initial temperature = 305 K
Final temperature = 32°C + 273.15 = 305.15 K
Initial pressure = 2 atm
final pressure = 101 kPa = 0.996792 atm
R = gasconstant = doesn't change
V1 = initial volume
V2= the final volume
<u>Step 2: </u>Calculate volume of original gas
P*V = n*R*T
(P*V)/ T = constante
(P1 * V1) / T1 = (P2 * V2)/ T2
In this situation we have:
(2atm * V1)/ 305 = (0.996792 *V2) / 305.15
0.006557*V1 = 0.003266*V2
V2 = 2*V1
We see that the final volume is twice the initial volume. So the volume gets doubled. The volume increases by 100%.
