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frozen [14]
2 years ago
11

Consider the reaction . At equilibrium, a 10 L vessel contains 1.5 moles of HI, 0.1 moles of I2 and 0.3 moles of H2. Which direc

tion will the reaction proceed if 2 moles of HI, and 0.5 moles of both I2 and H2 are mixed in a separate 10 L vessel
Chemistry
1 answer:
tankabanditka [31]2 years ago
5 0

The reaction will proceed in the forward direction if 2 moles of HI and 0.5 moles of both I_2  and H_2  are mixed in a separate 10 L vessel.

<h3>What is equilibrium?</h3>

In chemistry, equilibrium is the condition existing when a chemical reaction and its reverse reaction occur at equal rates.

If K > Q, a reaction will proceed forward, converting reactants into products. If K < Q, the reaction will proceed in the reverse direction, converting products into reactants. If Q = K then the system is already at equilibrium.

In the first case K < Q, but in the second case K > Q, so the reaction will proceed forward in the forward direction.

Learn more about the equilibrium here:

brainly.com/question/13463225

#SPJ1

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650. J is the same amount of energy as A. 155 cal. B. 2720 calC. 650 cal.D. 1550 cal.
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Answer:

A\text{ : 155 cal}

Explanation:

Here, we want to convert J to cal

Mathematically:

1\text{ cal = 4.186 J}

Thus, to get our answer in cal, we divide the value given by 4.186

We have that as:

\frac{650}{4.186}\text{ = 155 cal}

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Protons goes in the blank. the word can be used for both.
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An inclined plane is a machine why?​
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3 years ago
Rank the members of each set of compounds according to the ionic character of their bonds. Most ionic bonds?a) PCl3 PBr3 PF3 Mos
SCORPION-xisa [38]

Explanation:

More is the electronegativity difference between the combining atoms more polar is the compound. Hence, more ionic it will be in nature.

(a)   Electronegativity value of P = 2.19

Electronegativity value of Cl = 3.16

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a P-Cl bond = 3.16 - 2.19 = 0.97

Electronegativity difference of a P-Br bond = 2.96 - 2.19 = 0.77

Electronegativity difference of a P-F bond = 3.98 - 2.19 = 1.79

Since, a P-F bond has the highest electronegativity difference. Therefore, PF_{3} is the most ionic compound and PBr_{3} is the least ionic compound.

(b)   Electronegativity value of B = 2.04

Electronegativity value of N = 3.04

Electronegativity value of C = 2.55

Electronegativity value of F = 3.98

Electronegativity difference of a B-F bond = 3.98 - 2.04 = 1.94

Electronegativity difference of a N-F bond = 3.04 - 3.98 = 0.94

Electronegativity difference of a C-F bond = 3.98 - 2.55 = 1.43

Since, a B-F bond has the highest electronegativity difference. Therefore, BF_{3} is the most ionic compound and NF_{3} is the least ionic compound.

(c)   Electronegativity value of Se = 2.55

Electronegativity value of Te = 2.1

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a Se-F bond = 3.98 - 2.55 = 1.43

Electronegativity difference of a Te-F bond = 3.98 - 2.1 = 1.88

Electronegativity difference of a Br-F bond = 3.98 - 2.19 = 1.02

Since, a Te-F bond has the highest electronegativity difference. Therefore, TeF_{4} is the most ionic compound and BrF_{3} is the least ionic compound.

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Answer:

KCI

Explanation:

I took the test

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