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DIA [1.3K]
3 years ago
14

A hydrogen halide diffuses 1.49 times faster than HBr. This hydrogen halide is

Chemistry
1 answer:
marysya [2.9K]3 years ago
6 0

To solve this problem, we must assume ideal gas behaviour so that we can use Graham’s law:

vA / vB = sqrt (MW_B / MW_A)

where,

<span>vA = speed of diffusion of A  (HBR)</span>

vB = speed of diffusion of B (unknown)

MW_B = molecular weight of B (unkown)

MW_A = molar weight of HBr = 80.91 amu

 

We know from the given that:

vA / vB = 1 / 1.49

 

So,

1/1.49 = sqrt (MW_B / 80.91)

MW_B = 36.44 g/mol

 

Since this unknown is also hydrogen halide, therefore this must be in the form of HX.

HX = 36.44 g/mol , therefore:

x = 35.44 g/mol

 

From the Periodic Table, Chlorine (Cl) has a molar mass of 35.44 g/mol. Therefore the hydrogen halide is:

HCl

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2 years ago
Draw an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. (Assume standard conditions).
babunello [35]

Answer:

-1.05 V

Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

Oxidation half equation;

Sn(s) ------> Sn^2+(aq) + 2e

Reduction half equation:

Mn^2+(aq) + 2e ----> Mn(s)

Cell voltage= E°cathode - E°anode

E°cathode= -1.19V

E°anode= -0.14 V

Cell voltage= -1.19 V - (-0.14V)

Cell voltage= -1.05 V

8 0
3 years ago
The temperature of a sample of water changes from 10°C to 20°C when the water absorbs 100 calories of heat. What is the mass of
Vlad1618 [11]

Answer:

10 g

Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be  

10 g

.

Now, here's what that is the case.

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of  

1 g

of that substance by  

1

∘

C

.

Water has a specific heat of approximately  

4.18

J

g

∘

C

. This tells you that in order to increase the temperature of  

1 g

of water by  

1

∘

C

, you need to provide  

4.18 J

of heat.

Now, how much heat would be required to increase the temperature of  

1 g

of water by  

10

∘

C

?

Well, you'd need  

4.18 J

to increase it by  

1

∘

C

, another  

4.18 J

to increase it by another  

1

∘

C

, and so on. This means that you'd need

4.18 J

×

10

=

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

.

Now look at the value given to you. If you need  

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

, what mass of water would require  

10

times as much heat to increase its temperature by  

10

∘

C

?

1 g

×

10

=

10 g

And that's your answer.

Mathematically, you can calculate this by using the equation

q

=

m

⋅

c

⋅

Δ

T

 

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

418

J

=

m

⋅

4.18

J

g

∘

C

⋅

(

20

−

10

)

∘

C

m

=

418

4.18

⋅

10

=

10 g

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SVETLANKA909090 [29]

Answer:

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Explanation:

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4 0
2 years ago
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