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3 years ago

Carbon is ideally suited to form ionic bonds. True or false.

1 answer:

5 0

It is a completely false statement that carbon <span>is ideally suited to form ionic bonds. The correct option among all the options that are given in the question is the second option. Carbon is ideally suited to form covalent bonds. I hope that this is the answer that has actually come to your desired help.</span>

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What are the purposes of a chemical equation? Check all that apply.

hjlf

The answer is the first one

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3 years ago

Read 2 more answers

Traspasa el aire todo hasta llegar a la más alta esfera, y oye allí otro modo de no perecedera música, que es la fuente y la pri

kakasveta [241]

No entiendo cual es tu pregunta?

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3 years ago

ANSWER QUICK PLEASE <br> WORTH 100 POINTS

sukhopar [10]

Answer:

Explanation:

Volume is defined as the space occupied by an object or substance irrespective of its state of matter.The conversion used from millimeter to liter is:

1 milliiliter = 0.001 L

Therefore, we can convert the volume of sample from 2.5 ml in liters as follows.

2.5 ml in liters = 2.5ml x 0.001 L/1ml

= 0.0025 L

Thus, we can conclude that the volume of given sample in liter is 0.0025 L

Hope this helps! :)

6 0

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

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

What is the density of a liquid if it's volume is 125 mL and it's mass is 50g?

masha68 [24]

Answer:

0.4g/ml

Explanation:

density= mass/volume

density=50g/125ml

density=0.4g/ml

6 0

2 years ago