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melomori [17]
3 years ago
12

I’LL MAKE YOU BRAINLIEST+ FREE POINTS

Chemistry
1 answer:
vladimir2022 [97]3 years ago
6 0

Answer:

The solution's new volume is 1.68 L

Explanation:

Dilution is the procedure to prepare a less concentrated solution from a more concentrated one, and simply consists of adding more solvent. So, in a dilution the amount of solute does not vary, but the volume of the solvent varies.

In summary, a dilution is a lower concentration solution than the original.

The way to do the calculations in a dilution is through the expression:

Ci*Vi=Cf*Vf

where C and V are concentration and volume, respectively; and the i and f subscripts indicate initial and final respectively.

In this case, being:

  • Ci= 7 M
  • Vi= 0.60 L
  • Cf= 2.5 M
  • Vf=?

Replacing:

7 M*0.60 L= 2.5 M* Vf

Solving:

Vf=\frac{7 M*0.60 L}{2.5 M}

Vf= 1.68 L

<u><em>The solution's new volume is 1.68 L</em></u>

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Compare the solubility of silver chromate in each of the following aqueous solutions: Clear All 0.10 M AgCH3COO 0.10 M Na2CrO4 0
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Solution :

Comparing the solubility of silver chromate for the solutions :

$0.10 \ M \ AgCH_3COO$    -----     Less soluble than in pure water.

$0.10 \ M \ Na_2CrO_4$   ----- Less soluble than in pure water.

$0.10 \ M \ NH_4NO_3$   -----   Similar solubility as in the pure water

$0.10 \ M \ KCH_3COO$   -----   Similar solubility as in the pure water

The silver chromate dissociates to form :

$AgCrO_4 (s) \rightleftharpoons 2Ag^+ (aq) +CrO_4^{2-}(aq)$

When 0.1 M of $AgCH_3COO^-$ is added, the equilibrium shifts towards the reverse direction due to the common ion effect of Ag^+, so the solubility of Ag_2CrO_4 decreases.

Both AgCH_3COO and $KCH_3COO$ are neutral mediums, so they do not affect the solubility.

 

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Bill's recipe for onion soup calls for 4.0 lb of thinly sliced onions . if an onion has an average mass of 115 g , how many onio
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BillBill's recipe for onion soup calls for 4.0 lb of thinly sliced onions . if an onion has an average mass of 115 g , how many onions does Bill need ?

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An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr
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<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa

Rate of flow of ideal gas , n = 4 kmol/hr = \frac{4\times 1000mol}{3600s}=1.11mol/s    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

We know that:

\Delta U=0   (For isothermal process)

So, by applying first law of thermodynamics:

\Delta U=\Delta q-\Delta W

\Delta q=\Delta W      .......(1)

Now, calculating the work done for isothermal process, we use the equation:

\Delta W=nRT\ln (\frac{P_1}{P_2})

where,

\Delta W = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

P_1 = initial pressure = 100 kPa

P_2 = final pressure = 50 kPa

Putting values in above equation, we get:

\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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