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3 years ago

Addition of 1 M HBr to 0.1 M solutions of which

Chemistry

1 answer:

fgiga [73]3 years ago

8 0

Addition of 1 M HBr to 0.1 M solutions of the compound (A) NaHSO3 results in evolution of a colorless gas.

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Does the motion of the moon affects how we see it from Earth? *

irina [24]

Yes that’s why we see it in different shapes all the time

7 0

2 years ago

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

8 l of water at 19 ∘c are mixed with 1.1 l of water at 47 ∘c in an insulated container. what is the final temperature of the sys

stepladder [879]

1692.9 is the answer

5 0

3 years ago

When hydrogen is combined with a metal, it has an oxidation number of _____.

solmaris [256]

The answer is c because thats the number when combined with a metal. +1 is when its combined with a nonmetal.

8 0

3 years ago

Draw the lewis structure of the hypochlorite ion clo- . include lone pairs

Dmitry_Shevchenko [17]

Solution:- Hypochlorite ion $ClO^-$ has one Cl and one O atom. Cl has 7 valence electrons and O has 6 valence electrons. Since there is one negative charge on the ion,

total valence electrons = 7 + 6 +1 = 14

(note:- if there is negative charge then it is added and if there is positive charge then it is subtracted while calculating the valence electrons)

Both Cl and O atoms wants to complete their octet and so for this we put a single bond between them. Single bond means two electrons, so the remaining electrons would be 14 - 2 = 12

It means 12 electrons will be placed as lone pair of electrons. To complete the octet, we put 6 dots around each of the atom. Oxygen is more electron negative than Cl, so we show the -1 charge for oxygen.

6 0

3 years ago