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posledela
1 year ago
7

How many grams of agcl would be needed to make a 4. 0 m solution with a volume of 0. 75 l?

Chemistry
1 answer:
devlian [24]1 year ago
8 0

430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

<h3>What is Molarity?</h3>
  • The amount of a substance in a specific volume of solution is known as its molarity (M).
  • The number of moles of a solute per liter of a solution is known as molarity.
<h3>Calculation of Required amount of AgCl</h3>

Remember that mol/L is the unit of molarity (M).

We can compute the necessary number of moles of solute by multiplying the concentration by the liters of solution, according to dimensional analysis.

0.75L×4.0M=3.0mol

Then, using the periodic table's molar mass for AgCl, convert from moles to grams:

3.0mol×143.321gmol=429.963g

The final step is to round to the correct significant figure, which in this case is two: 430g.

Hence, 430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

Learn more about Molarity here:
brainly.com/question/8732513

#SPJ4

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Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

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Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

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<h2>The answer is option A</h2>

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