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Rudik [331]
3 years ago
10

As you pass a magnet over a blackish-yellow powder on a glass plate, black particles separate from the powder and become attache

d to the magnet. After a few passes of the magnet over the powder nothing but a yellow powder remains in the glass plate. The original blackish-yellow powder was:
Chemistry
1 answer:
finlep [7]3 years ago
4 0

AS given that we have started with a solid powder on a glass plate which appears to be blackish yellow, on passing magnet the black particles separated.

so the black particles are magentic in nature. however the yellow portion of powder is not magnetic or diamagnetic

so its clear that we are able to separate the black and yellow portions by simple physical methods.

If we are able to separate any two substances by simple physical methods then it means that the powder is a mixture and not a compount

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The rate constant for the decomposition reaction of H2O2 is 3.66 × 10−3 s−1 at a particular temperature. What is the concentrati
Pepsi [2]

Answer:  3.72 M

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant = 3.66\times 10^{-3}s^{-1}

t = age of sample = 15.0 minutes

a = let initial amount of the reactant  = 10.0 M

a - x = amount left after decay process = ?

15.0\times 60s=\frac{2.303}{3.66\times 10^{-3}}\log\frac{10.0}{(a-x)}

\log\frac{100}{(a-x)}=1.43

\frac{100}{(a-x)}=26.9

(a-x)=3.72M

The concentration of H_2O_2 in a solution after 15.0 minutes have passed is 3.72 M

7 0
3 years ago
Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p
UNO [17]

Answer : The equilibrium constant K_c for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of Br_2.

\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}

\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M

Now we have to calculate the dissociated concentration of Br_2.

The balanced equilibrium reaction is,

                              Br_2(g)\rightleftharpoons 2Br(aq)

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of Br_2 = \alpha = 1.2 %

So, the dissociate concentration of Br_2 = C\alpha=1.731M\times \frac{1.2}{100}=0.2077M

The value of x = 0.2077 M

Now we have to calculate the concentration of Br_2\text{ and }Br at equilibrium.

Concentration of Br_2 = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of Br = 2x = 2 × 0.2077 = 0.4154 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{[Br]^2}{[Br_2]}

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

K_c=\frac{(0.4154)^2}{1.5233}=0.1133

Therefore, the equilibrium constant K_c for the reaction is, 0.1133

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