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enot [183]
3 years ago
8

The circuit element whose purpose is to convert electrical energy into another form of energy is the _____. battery resistor fus

e ammeter
Chemistry
2 answers:
ludmilkaskok [199]3 years ago
8 0

Answer : The correct option is, Resistor.

Explanation :

Battery : It is a type of device that contains one or more cells in which the chemical energy converted into electrical energy. The battery is used as a source of power.

Resistor : It is a device which resist's the flow of current in an electric circuit. When an electric current pass through a resistor it gets heated up due to the conversion of electrical energy into heat energy.

Fuse : It is a safety device which contains a strip of wire that breaks an electric current when the current increases to a safe level.

Ammeter : It is an electrical device in which measures the electric current in amperes.

Hence, the correct option is, Resistor.

svetoff [14.1K]3 years ago
6 0
The answer is Battery...
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Carry out the following operations as if they were calculations of experimental results and express each answer in standard nota
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Answer:

1. 10.6\; \rm m (one decimal place.)

2.0.79\; \rm g (two decimal places.)

3. 16.5\;\rm cm^2 (three significant figures.)

Explanation:

<h3>1.</h3>

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

  • 5.6792\;\rm m has four decimal places.
  • 0.6\; \rm m has only one decimal place.
  • 4.33\; \rm m has two decimal places.

Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: \rm m.)

Therefore:

\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m. (Rounded to one decimal place.)

<h3>2.</h3>

Similarly:

  • \rm 3.70\; \rm g has two decimal places.
  • 2.9133\; \rm g has four decimal places.

Therefore, the result should be rounded to two decimal places. Its unit should be \rm g (same as the unit of the two inputs.)

\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g. (Rounded to two decimal places.)

<h3>3.</h3>

When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:

  • 4.51\; \rm cm has three significant figures.
  • 3.6666\; \rm cm has five significant figures.

Therefore, the result should have only three significant figures.

The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be \rm cm \cdot cm, which is occasionally written as \rm cm^2.

\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2. (Rounded to three significant figures.)

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3 years ago
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3 years ago
A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express
Rudik [331]

Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

(c) Molarity = 13792 mol HNO3/m3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\  \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3

(c) HNO3 molarity (mol HNO3/m3)

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are  1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3

We can now calculate the molarity as

Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3}  =13792 \frac{molHNO3}{m3}

8 0
3 years ago
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