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WARRIOR [948]
2 years ago
15

Suppose you have a system made up of water only, with the container and everything beyond being the surroundings. Consider a pro

cess in which the water is first evaporated and then condensed back into its original container.
Is this two-step process necessarily reversible?
Chemistry
2 answers:
Airida [17]2 years ago
8 0

Answer:

Yes

Explanation:

The possibility of evaporating and condensing is a proof of reversible reaction

natima [27]2 years ago
6 0

Answer:

No

Explanation:

No, because there is no evidence that the surroundings are also restored to their original state.

cannot be undone by exactly reversing the change to the system; Spontaneous processes are irreversible.

You might be interested in
II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo
Leto [7]

Answer:

Excess=3.53g

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2Mg(s)+O_2(g) \rightarrow 2MgO(s)

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg}  =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg

Thus, the mass in excess is:

Excess=3.86g-0.334g\\\\Excess=3.53g

Regards!

5 0
2 years ago
A heat energy of 645 J is applied to a sample of glass with a mass of 28.4 g. Its temperature increases from –11.6 ∞C to 15.5 ∞C
Anika [276]
The heat that is required to raise the temperature of an object is calculated through the equation,
                        heat = mass x specific heat x (T2 - T1)
Specific heat is therefore calculated through the equation below,
                                specific heat = heat / (mass x (T2 - T1))
Substituting,
                                specific heat = 645 J / ((28.4 g)(15.5 - - 11.6))
The value of specific heat from above equation is 0.838 J/g°C. 
5 0
2 years ago
Iodine-131 has a half-life of 8.10 days. In how many days will 50 grams of Iodine-131 decay to one-eighth of its original amount
shtirl [24]
What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half 
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25                                               8.1
25/2 =12.5                                        +  8.1
12.5/2= 6.25                                      +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8  of its original mass.

(good luck on the regent if thats what your studying for :)

5 0
2 years ago
Read 2 more answers
At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.400 M . N 2 ( g )
stiks02 [169]

<u>Answer:</u> The equilibrium concentration of NO after it is re-established is 0.55 M

<u>Explanation:</u>

For the given chemical equation:

N_2(g)+O_(g)\rightleftharpoons 2NO(g)

The expression of K_{eq} for above equation follows:

K_{eq}=\frac{[NO]^2}{[N_2][O_2]}     .....(1)

We are given:

[NO]_{eq}=0.400M

[N_2]_{eq}=0.200M

[O_2]_{eq}=0.200M

Putting values in expression 1, we get:

K_{eq}=\frac{(0.400)^2}{0.200\times 0.200}\\\\K_{eq}=4

Now, the concentration of NO is added and is made to 0.700 M

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

The equilibrium will shift in backward direction.

                           N_2(g)+O_(g)\rightleftharpoons 2NO(g)

<u>Initial:</u>               0.200    0.200        0.700

<u>At eqllm:</u>      0.200+x   0.200+x     0.700-2x

Putting values in expression 1, we get:

4=\frac{(0.700-2x)^2}{(0.200+x)\times (0.200+x)}\\\\x=0.075

So, equilibrium concentration of NO after it is re-established = (0.700 - 2x) = [0.700 - 2(0.075)] = 0.55 M

Hence, the equilibrium concentration of NO after it is re-established is 0.55 M

4 0
2 years ago
If the ph of a solution is decreased from ph 8 to ph 6 it means that the
saw5 [17]

Explanation:

Relation between pH and concentration of hydrogen ions is as follows.

                  pH = -log [H^{+}]

So, it means that an increase in the value of pH will show that there occurs a decrease in concentration of hydrogen ions.

Therefore, the solution becomes basic in nature.

On the other hand, a decrease in the value of pH will show that there occurs an increase in the concentration of hydrogen ions.

Therefore, the solution becomes more acidic in nature.

Hence, if the pH of a solution is decreased from pH 8 to pH 6 it means that the concentration of hydrogen ions has increased in the solution.

5 0
2 years ago
Read 2 more answers
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