The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer:
Explanation:
MM of I2 = 2 (127 g) = 254 g/mol
0.065 mol I2 x 254g I₂/ 1 mol I₂ = 16.5 g I2
Answer:
17 ohms
Explanation:
Given that,
In a series circuit there are 17 light bulbs, each providing 1 ohm of resistance.
We need to find the total resistance of the circuit.
The equivalent resistance for series combination is given by :
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For 17 light bulbs, the equivalent resistance is given by :
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Hence, the total resistance is 17 ohms. Hence, the correct option is (d).
The full question is shown in the image attached
Answer:
See explanation
Explanation:
In naming an alkane, the first thing we do is to obtain the parent chain by counting the number of carbon atoms in the chain.
When we obtain that, then we identify the substituents and number them in such a way that they have the lowest numbers. The compounds shown have the following names according to the order in which the structures appear in the image attached;
1. 2-methyl propane
2. 2,4-dimethyl heptane
3. 2,2,3,3-tetramethyl butane
4. 5-ethyl-2,4-dimethyl octane
Answer:
1.7 bar
Explanation:
We can use the <em>Ideal Gas Law</em> to calculate the individual gas pressure.
pV = nRT Divide both sides by V
p = (nRT)/V
Data: n = 1.7 × 10⁶ mol
R = 0.083 14 bar·L·K⁻¹mol⁻¹
T = 22 °C
V = 2.5 × 10⁷ L
Calculations:
(a) <em>Change the temperature to kelvins
</em>
T = (22 + 273.15) K
= 295.15 K
(b) Calculate the pressure
p = (1.7 × 10⁶ × 0.083 14 × 295.15)/(2.5× 10⁷)
= 1.7 bar