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3 years ago

Need some help please

Chemistry

1 answer:

Marysya12 [62]3 years ago

5 0

Number 3:
Chlorine, Sodium, Sulfate, Magnesium, and Calcium.
Number 4:
The salt Increases/Decreases the density.
Hope this helps you!
:)

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Volume of Water -

miss Akunina [59]

Answer:

3.8 g/mL

Explanation:

From the question given above, the following data were obtained:

Volume of Water = 60 mL

Volume of Water + Object = 73.5 mL

Mass of object = 51.3 g

Density of object =?

Next, we shall determine the volume of the object. This can be obtained as follow:

Volume of Water = 60 mL

Volume of Water + Object = 73.5 mL

Volume of object =?

Volume of object = (Volume of Water + Object) – (Volume of Water)

Volume of object = 73.5 – 60

Volume of object = 13.5 mL

Finally, we shall determine the density of the object as illustrated below:

Volume of object = 13.5 mL

Mass of object = 51.3 g

Density of object =?

Density = mass /volume

Density of object = 51.3 /13.5

Density of object = 3.8 g/mL

Thus, the density of the object is 3.8 g/mL

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3 years ago

What is the total number of molecules in 34.0?

mr Goodwill [35]

What is the element in which you are trying to find the number of molecules of?

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3 years ago

The electrolyte magnesium chloride (MgCl2) will break up into how many individual particles in water (what is the i value)?

Masja [62]

Answer:

Solid magnesium chloride is a non-conductor of electricity because the ions aren't free to move. However, it undergoes electrolysis when the ions become free on melting. Magnesium chloride dissolves in water to give a faintly acidic solution (pH = or 6).

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3 years ago

Round 1532.2364 to 3 significant figures.

mote1985 [20]

Answer:

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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.

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4 years ago