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Andrej [43]
3 years ago
10

Two children bring home a salt-water fish for their new fish tank. The children are eager to put their fish into the new fish ta

nk, but have to
create a salt water solution to fill the tank first. The children plan to compete to see whose solution will be finished first.
Particle Size of Salt
Temperature of Solvent
Level of Agitation
Solution A
Sea salt (small clumps)
Cold water
Slow stirring
Solution B
Granular salt (tiny grains)
Hot water
Fast stirring
What do you believe the outcome will be from these solutions?
O A Solution A will take less time to dissolve, because gently stirring will allow it to combine more evenly
OB Solution will take less time to dissolve, because hot water will cause some of the salt to evaporate.
O c Solution A will take longer to dissolve and might have some undissolved salt remaining at the bottom of the solution due to the low
temperature.
OD Solution will take longer to dissolve and might have some undissolved salt remaining at the bottom of the solution due to the
high temperature

Chemistry
1 answer:
Elena-2011 [213]3 years ago
5 0

Answer:

b.

Explanation:

. i got it from my smart friend

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An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
3 years ago
if i mix blue gatorade with red gatorade what do i get this is for my science exam right now please answer
Gnesinka [82]

Answer:

purple gatorade

Explanation:

:)

7 0
2 years ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
How many neutrons does Ironatomhave?
luda_lava [24]
The answer would be 26!
5 0
3 years ago
Read 2 more answers
A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:2NH3(g) + 3I2(g) ⇌ N2(g) + 6
miss Akunina [59]

Answer: The value of K_{c} for this reaction is 250000.

Explanation:

The given equation is as follows.

2NH_{3}(g) + 3I_{2}(g) \rightleftharpoons N_{2}(g) + 6HI(g)

N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g); K_{c_{1}} = 0.50   ... (1)

H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g); K_{c_{2}} = 50  ... (2)

To balance the atoms, multiply equation (2) by 3. Hence, the equation (2) can be re-written as follows.

3H_{2}(g) + 3I_{2}(g) \rightleftharpoons 6HI(g); K_{c_{2}} = (50)^{3}  ... (3)

Now, subtract equation (1) from equation (3). So, the equation formed will be as follows.

3I_{2} - N_{2} \rightleftharpoons 6HI - 2NH_{3}

This equation can also be re-written as follows.

3I_{2} + 2NH_{3} \rightleftharpoons N_{2} + 6HI

This equation is similar to the equilibrium equation given to us.

Therefore, during this subtraction the equation constants get divided as follows.

K^{'}_{c} = \frac{K_{c_{2}}}{K_{c_{1}}}\\= \frac{(50)^{3}}{0.50}\\= 250000

Thus, we can conclude that the value of K_{c} for this reaction is 250000.

6 0
2 years ago
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