Ask question

Ask question

All categories

3 years ago

5

How can you force a solid compound to the bottom of a melting point capillary? a. tap the closed endb. tap the open endc. use a

needle

1 answer:

snow_lady [41]3 years ago

3 0

Answer: option A. tap the closed end

Explanation:

You might be interested in

Which state of matter has the MOST energy?

Ierofanga [76]

The state of matter that has the most energy is gas. In a solid, there is limited room for molecules to move around.

The gas state of matter has the most energy because of how freely the molecules move. Matter is a physical substance of which there are three, and they are solid, liquid, and gas. Solid matter is very compact with the particles very close together and not much movement; therefore, no action, no energy.

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em>

4 0

3 years ago

Read 2 more answers

.325 mol KOH to grams

lesya [120]

The answer is 18.23432 grams.

Molar mass of KOH= 56.1056

$( \frac{1 mol}{56.1056 grams}) (\frac{.325 mol}{x grams})$ = 18.234 grams

5 0

3 years ago

Helen recorded the following data about the half-life of a radioisotope. Radioactive Decay of Radioisotope A Grams of Radioactiv

MissTica

Answer:

Line graph

Explanation:

I did it on Study Island

7 0

4 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

4 years ago

What causes a sharp increase in potential energy when two

Pachacha [2.7K]

The answer is c electrostatic repulsion between nuclei

4 0

3 years ago

Read 2 more answers