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4 years ago

How is solute concentration related to the solute-solvent interactions and the rate at which the solute dissolves?

Chemistry

2 answers:

shusha [124]4 years ago

8 0

Answer:

Explanation:

solute is what gets dissolved in the solvent

ie: salt in water. salt is the salute and the more of it you add to the water (solvent) the more interactions will occur increasing rate

Troyanec [42]4 years ago

8 0

Answer: B.

Explanation:

Increasing the solute concentration increases the solute-solvent interactions and increases the rate at which the solute dissolves.

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3 years ago

Consider the general reaction: 2 A + b B → c C and the following average rate data over a specific time period \Delta t: - \frac

lawyer [7]

Answer:

c = 4

Explanation:

In general, for the reaction

a A + b B ⇒ c C + d D

the rate is given by:

rate = - 1/a ΔA/Δt = - 1/b ΔB/Δt = + 1/c ΔC/Δt = + 1/d ΔD/Δt

this is done so as to express the rate in a standarized way which is the same to all the reactants and products irrespective of their stoichiometric coefficients.

For this question in particular we know the coefficient of A and need to determine the coefficient c.

- 1/2 ΔA/Δt = + 1/c ΔC/Δt

- 1/2 (-0.0080 ) = + 1/c ( 0.0160 mol L⁻¹s⁻¹ )

0.0040 mol L⁻¹s⁻¹ c = 0.0160 mol L⁻¹s⁻¹

∴ c = 0.0160 / 0.0040 = 4

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3 years ago

While camping up at Haloooo Mountain (5,500 feet above sea level), it generally takes Sandy about the same amount of time to boi

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She should make soup. When you are way above sea level, the atmospheric pressure decreases. Boiling starts when the vapor pressure of the liquid is the same with the atmospheric pressure. This would mean that water boils at a much lower temperature than the usual 100deg C. So it would take longer for the components in the soup to cook compared to when it is cooked on the foot of the mountain.

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3 years ago

Nitrogen (N2) enters a well-insulated diffuser operating at steady state at 0.656 bar, 300 K with a velocity of 282 m/s. The inl

Tasya [4]

Answer:

The exit temperature of the Nitrogen would be 331.4 K.
The area at the exit of the diffuser would be $7*10^{-3} m^2$ .
The rate of entropy production would be 0.

Explanation:

First it is assumed that the diffuser works as a isentropic device. A isentropic device is such that the entropy at the inlet is equal that the entropy T the exit.
It will be used the subscript 1 for the inlet conditions of the nitrogen, and the subscript 2 for the exit conditions of the nitrogen.
It will be called: v the velocity of the nitrogen stream, T the nitrogen temperature, V the volumetric flow of the specific stream, A the area at the inlet or exit of the diffuser and, P the pressure of the nitrogen flow.
It is known that for a fluid flowing, its volumetric flow is obtain as: $V=v*A$ ,
Then for the inlet of the diffuser: $V_1=v_1*A_1=282\frac{m}{s}*4.8*10^{-3}m^2=1.35\frac{m^3}{s}$
For an ideal gas working in an isentropic process, it follows that: $\frac{T_{2} }{T_1}=(\frac{P_2}{P_1})^k$ where each variable is defined according with what was presented in step 2 and 3, and k is the heat values relationship, 1.4 for nitrogen.
Then solving for $T_2$ , the temperature of the nitrogen at the exit conditions: $T_2=T_1(\frac{P_2}{P_1})^k$ then, $T_2=300 K (\frac{0.9 bar}{0.656 bar})^{(\frac{1.4-1}{1.4})}=331.4 K$
Also, for an ideal gas working in an isentropic process, it follows that: $\frac{P_2}{P_1}= (\frac{V_1}{V_2})^k$ , where each variable is defined according with what was presented in step 2 and 3, and k is the heat values relationship, 1.4 for nitrogen.
Then solving for $V_2$ the volumetric flow at the exit of the diffuser: $V_2=V_1*\frac{1}{\sqrt[k]{\frac{P_2}{P_1}}}=\frac{1.35\frac{m^3}{s}}{\sqrt[1.4]{\frac{0.9bar}{0.656bar} }}=1.080\frac{m^3}{s}$ .
Knowing that $V_2=1.080\frac{m^3}{s}$ , it is possible to calculate the area at the exit of the diffuser, using the relationship presented in step 4, and solving for the required parameter: $A_2=\frac{V_2}{v_2}=\frac{1.08\frac{m^3}{s} }{140\frac{m}{s}}=7.71*10^{-3}m^2$ .
To determine the rate of entropy production in the diffuser, it is required to do a second law balance (entropy balance) in the control volume of the device. This balance is: $S_1+S_{gen}-S_2=\Delta S_{system}$ , where: $S_1$ and $S_2$ are the entropy of the stream entering and leaving the control volume respectively, $S_{gen}$ is the rate of entropy production and, $\Delta S_{system}$ is the change of entropy of the system.
If the diffuser is operating at stable state is assumed then $\Delta S_{system}=0$ . Applying the entropy balance and solving the rate of entropy generation: $S_{gen}=S_2-S_1$ .
Finally, it was assume that the process is isentropic, it is: $S_1=S_2$ , then $S_{gen}=0$ .

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3 years ago