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3 years ago

How many moles H2SO4 are needed to make 85 mL of .750 M solution?

Chemistry

1 answer:

grin007 [14]3 years ago

4 0

Answer:

0.06375 mole

Explanation:

The following data were obtained from the question:

Volume = 85mL

Molarity of H2SO4 = 0.750 M

Mole of H2SO4 =..?

Next, we'll convert 85mL to a value in L.

This is illustrated below:

1000mL = 1L

Therefore, 85mL = 85/1000 = 0.085L

Now, we can obtain the number of mole of H2SO4 as follow:

Molarity is simply defined as the mole of solute per unit litre of the solution i.e

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of H2SO4 = 0.750 x 0.085

Mole of H2SO4 = 0.06375 mole

Therefore 0.06375 mole of H2SO4 is needed to make the solution

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Describe polar covalent bonds using water as an example

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2 years ago

Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int

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Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.

Million Gallons per day $1 MGD = 3785411.8 litre/day = 3785411.8 L/d$

$F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d$

We have one flow of wastewater released into a stream.

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:

$C_f = \frac{F1*C1 +F2*C2}{F1 +F2}$

Replacing every value in L/d and mg/L

$C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L$

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).

We have the total flow F3 = F1 + F2 and the final concentration $C_f$ . It is required to calculate per day, let's take a time of t = 1 day.

$F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg$

After that, mg are converted to pounds.

$M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb$

b) A total of 137.6 pounds pass a given spot downstream per day.

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3 years ago

What will be the answee for the quistion <br>solve for x 2x=1

Nadusha1986 [10]

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Answer:

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