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3 years ago

11

In the following diagram, the voltage is 1.5 volts and the resistance is 7.5 ohms. Use Ohm's law to determine the current in the

circuit.

2 answers:

Alona [7]3 years ago

6 0

<u>Answer:</u> The current flowing through the circuit is 0.2 Amperes.

<u>Explanation:</u>

Ohm's Law is defined as the law which determines the relationship between current and potential difference across a resistor.

Mathematically,

$V=IR$

where,

V = Potential difference across the resistor = 1.5 V

I = Current flowing through the circuit = ? A

R = Resistor in the circuit = $7.5\Omega$

Putting values in above equation, we get:

$1.5=I\times 7.5\\\\I=0.2A$

Hence, the current flowing through the circuit is 0.2 Amperes.

maks197457 [2]3 years ago

5 0

I believe it is 0.2 amps

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Determine the molarity for each of the following solution solutions:

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Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

<u>(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

<u>(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

<u>(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

<u>(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

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3 years ago

erastova [34]

Answer:

Problem: The density of ethanol is 0.789 g/cm3 .What is its density in pounds per cubic inch ( lb/in3)?. Based on our data, ... 1 in = 2.54 cm. density = 0 . 789 g ...

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Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a

s2008m [1.1K]

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Explanation:

Partial pressure of the $O_2$ gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

$p_{o_2}=K_H\times\chi_{O_2}$

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

$0.0013=34860 bar\times \chi_{O_2}$

$\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}$

Let the number of moles of $O_2$ gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L $n_w=\frac{1000 g}{18 g/mol}=55.55 mol$

$\chi_{O_2}=\frac{n}{n+n_w}$

$2.56\times 10^{-5}=\frac{n}{n+55.55}$

$n=1.43\times 10^{-7} moles$

$Molarity=\frac{\text{Moles of}O_2}{Volume}$

Molar concentration of oxygen gas in 1 L of water:

$=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L$

The molar concentration of oxygen gas in water is $1.43\times 10^{-7} mol/L$ .

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