What element would have this electron configuration:<br> 1s22s22p 3s 3p 4s 3d

lesya692 [45]

Answer:

B. Spring balance - a device used for measuring the weight or force of gravity acting on an object.

Explanation:

A Force is any interaction that changes the motion or position of an obkpjectbthatbit is interacting with. Whenever there is an interaction between two objects, there is a force exerted by each of the objects on one themselves.

Forces are generally divided into contact forces and non-contact over field forces.

In contact forces, the two objects physically in contact with each other. Examples of contact forces are push or pull forces, frictional forces, tensional forces, spring forces, etc.

Non-contact forces are forces in which the two objects interacting do no need to be physically in contact with one another. Examples include, gravitational forces, magnetic forces, electrical forces, etc.

Instruments used in measuring forces are known as force gauges.

From the instruments listed above:

A. A ruler is an instrument used in measuring length

B. Spring balance is a device used for measuring the weight or force of gravity acting on an object.

C. A thermometer is an instrument used in measuring temperature

D. A windbvane is an instrument used in measuring wind direction.

4 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

How many grams of iodine are needed to prepare 28.6 grams of ICl

Maslowich

The answer will be 17.9 grams

4 0

3 years ago

How many grams of barium sulfate, BaSO 4 , be precipitated when 2.25 moles of sodium sulfate , Na 2 SO 3 , reacts with an excess

gulaghasi [49]

Answer:

525.1 g of BaSO₄ are produced.

Explanation:

The reaction of precipitation is:

Na₂SO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) ↓ + 2NaCl (aq)

Ratio is 1:1. So 1 mol of sodium sulfate can make precipitate 1 mol of barium sulfate.

The excersise determines that the excess is the BaCl₂.

After the reaction goes complete and, at 100 % yield reaction, 2.25 moles of BaSO₄ are produced.

We convert the moles to mass: 2.25 mol . 233.38 g/mol = 525.1 g

The precipitation's equilibrium is:

SO₄⁻² (aq) + Ba²⁺ (aq) ⇄ BaSO₄ (s) ↓ Kps

7 0

2 years ago

How much energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°c and completely vaporize the sample?

natta225 [31]

The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.

Calculation,

Given data,

Mass of the ice = 10 g

Temperature of ice = 0. 0°C

The ice at 0. 0°C is to be converted into water at 0. 0°C

Heat required at this stage = mas of the ice ×latent heat of fusion of ice

Heat required at this stage = 10 g×80 = 800 cal

The temperature of the water is to be increased from 0. 0°C to 100. 0°C

Heat required for this = mass of the ice×rise in temperature×specific heat of water

Heat required for this = 10 g×100× 1 = 1000 cal

This water at 100. 0°C is to be converted into vapor.

Heat required for this = Mass of water× latent heat

Heat required for this = 10g ×536 =5360 cal

Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal

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6 0

1 year ago

What element would have this electron configuration: 1s22s22p 3s 3p 4s 3d