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Given the following information: benzoic acid = C6H5COOH hydrocyanic acid = HCN C6H5COOH is a stronger acid than HCN (1) Write t

Maurinko [17]

Answer:

The net ionic equation is

C6H5COOH+ CN-= C6H5COO- + HCN

Explanation:

From the ionic equation

C6H5COOH + Na+ + CN- = C6H5COO- + Na+ + HCN

Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated

5 0

3 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

hjlf

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.

4 0

3 years ago

Name the type of equation shown here: 2NaBr+Ca(OH)2 —>CaBr2+2NaOH

sergeinik [125]

A) Double replacement
Because the reactants switch when they become products

6 0

3 years ago

Read 2 more answers

How many control(s) are in an experiment

weqwewe [10]

You can have as many controls as necessary, But they must remain equal at all times in order to get the most accurate results

6 0

3 years ago

Acetic acid, ethanol, acetaldehyde and ethane form a series of 2-carbon molecules which differ in their extent of oxidation. Con

yanalaym [24]

Explanation:

Ethanol can be oxidized to ethanal or acetaldehyde which is further oxidized to acid that is acetic acid.

$CH_3CH_2OH$ → $CH_3CHO$ [oxidation by loss of hydrogen]

An oxidizing agent potassium dichromate(VI) solution is used to remove the hydrogen from the ethanol.

An oxidizing agent used along with dilute sulphuric acid for acidification.

Acetaldehyde can also be reduced back to ethanol again by adding hydrogen to it by using a reducing agent that is sodium tetrahydro borate, NaBH4.

The oxidation of aldehydes to carboxylic acids can be done by the two-step process.

In the first step, one molecule of water is added in the presence of a catalyst that is acidic.

There is a generation of a hydrate. (geminal 1,1-diol).

$CH_3CHO$ → $CH_3CH_2COOH$ [reduction by the gain of electrons]

Here, the oxidizing agent used is $Cr_3O$ in the presence of acetone.

7 0

3 years ago