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2 years ago

Write a balanced half-reaction describing the reduction of gaseous dibromine to aqueous bromide anions.

Chemistry

1 answer:

stich3 [128]2 years ago

3 0

<u>Answer:</u> The balanced half reaction is written below.

<u>Explanation:</u>

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

When bromine gas is reduced to bromide ions, the number of electron transferred are 2

The chemical equation for the reduction of bromine gas to bromide ions follows:

$Br_2+2e^-\rightarrow 2Br^{-}$

Hence, the balanced half reaction is written above.

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

8 0

2 years ago

We have an object with a density of 620 g/ cm3 and a volume of 75 cm3. What is the mass of this object?

Anna11 [10]

M=D*V
D=620 g/cm³
V=75 cm³

m= 620 g/cm³ * 75 cm³=46500 g
m=46500g

8 0

2 years ago

A student assistant is cleaning up after a chemistry laboratory exercise and finds three one-liter bottles containing alcohol so

Svetllana [295]

The approximate alcohol content is 210 ml.

Explanation:

It can be deduced from the question that each bottle is of 1000ml or 1 litre.

The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is

20/100*500

=100 ml

The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml

so it is 200ml having 30% alcohol

30/100*200

= 60 ml

The third bottle is one tenth full so its volume is 1/10*1000

100 ml. having 50% of alcohol

50/100*100

50 ml.

The alcohol content obtained from all these 3 litres is:

100+60+50

= 210 ml of alchohol is obtained from 800 ml of mixture.

3 0

2 years ago

There are ________ hydrogen atoms in 25 molecules of c4h4s2.

coldgirl [10]

Answer: 100.

Explanation:

1) The subscripts to the right of each element (symbol) in the chemical formula tells the number of atoms of that element present in one unit formula.

2) The unit formula of C₄H₄S₂ is equal to 1 molecule.

3) Therefore, there are 4 carbon atoms, 4 hydrogen atoms and 2 sulfur atoms in each molecule of C₄H₄S₂.

4) Then, you just have to multiply the corresponding subscript of the element times the number of molecules (25 in this case) to find the number of atoms of that kind.

5) These are the calculations for each element in the molecule C₄H₄S₂.

i) C: 4 × 25 = 100

ii) H: 4 × 25 = 100

iii) S: 2 × 25 = 50.

6) The question is about H only, so the answer is that there are 100 hydrogen atoms in 25 molecules of C₄H₄S₂.

5 0

2 years ago

Read 2 more answers

In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Ke

kondor19780726 [428]

Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

$q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}$

$q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}$

so,

$q$ ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq

4 0

2 years ago