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3 years ago

Consider the reaction below.

Chemistry

1 answer:

patriot [66]3 years ago

6 0

Answer:

HI and I-

Explanation:

Hello there!

In this case, according to the acid ionization of hydroiodic acid:

$HI+H_2O\rightarrow I^-+H_3O^+$

Now, since the acid, HI, results in the conjugate base, I- and the base, H2O in the conjugate acid, H3O+, it is correct to assert that the acid-conjugate base pair is HI and I-

Best regards!

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Answer:

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Explanation:

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3 years ago

What is an acid base indicator

zhenek [66]

Answer:

Acid-base indicators are generally weak proteolytic that change color in solution according to the pH. The acid-base equilibrium of a weak acid type of indicator (HI) in water can be represented as. [I] The acid, HI, and the conjugate base, I−, have different colors. The equilibrium expression for this process is.

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3 years ago

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for

tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. $K(s)+O_2(g)\rightarrow KO_2(s)$

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$

Answer: A. $K(s)+O_2(g)\rightarrow KO_2(s)$ : decreases

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g)$ : decreases

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ : no change

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction $K(s)+O_2(g)\rightarrow KO_2(s)$ , as gaseous reactant is changed to solid product, entropy decreases.

In reaction $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ , as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ , as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

7 0

3 years ago

Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c

uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

$I_{2} (g) + Cl_{2} (g) -----> 2 ICl (g)$

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

$K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}$

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = $\frac{(2x)^{2}}{(0.437-x) (0.269-x)}$

81.9 = $\frac{(2x)^{2}}{x^{2} -0.706x + 0.118}$

$(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]$

$4x^{2} = 81.9x^{2} -57.8x +9.66$

$77.9x^{2} -57.8x+9.66 = 0$

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

$[ICl]_{eq} = 2 ( 0.254) = 0.508 M$

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

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3 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

4 years ago