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2 years ago

What is the oxidation state of an individual bromine atom in kbro2?

1 answer:

6 0

The oxidation state of potassium ion K = +1
The oxidation state of oxygen ion O = -2
So, the oxidation state of O2 is = -2 x 2 = -4
Since, KBrO2 is neutral so,
(+1) + (x) + (-4) = Zero
-3 + X = Zero
So, X = +3
The oxidation state of individual bromine atom in KBrO2 is +3

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1.05 Quiz: Measure Angles

Lelechka [254]

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2 years ago

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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

analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula

MrMuchimi

1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr /0.0135= 1 mol Cr
0.0538 mol O/0.035= 4 mol Cr
K2CrO4

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2 years ago

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What is the mass, in g, of 0.500 mol of 1,2-dibromoethane, CH2BrCH2Br?

inna [77]

Molar mass <span>CH2BrCH2Br = 188.0 g/mol

1 mole ---------- 188.0 g
</span>0.500 moles ----- ?

mass = 0.500 * 188.0 / 1

= 94.0 g

Answer C

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2 years ago

A scuba tank contains a mixture of oxygen (O2) and nitrogen (N2) gas. The oxygen has a partial pressure of PO2=5.62MPa. The tota

dmitriy555 [2]

Answer:

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Explanation:

Partial pressure of oxygen = 5.62 MPa

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Total pressure of the gas= sum of partial pressures of all the constituent gases in the system.

This implies that;

Total pressure of the system = partial pressure of nitrogen + partial pressure of oxygen

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Partial pressure of nitrogen= 26.78 - 5.62

Partial pressure of nitrogen = 21.16 MPa

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2 years ago

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