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tatiyna
3 years ago
15

When you turn on a battery-operated device, an electrical circuit is . opened closed

Chemistry
2 answers:
BlackZzzverrR [31]3 years ago
7 0

The answer is closed.

Explanation:

The circuit is now active because the key has been closed so electricity is flowing freely.

Charra [1.4K]3 years ago
4 0

Answer:

Closed

Explanation:

When the battery-operated device is of this means no current is flowing through the circuit. Since no current is flowing the device won't work. But since the battery is installed in the device and the current is not flowing this indicates that the circuit is not complete which is preventing the current from flowing in the circuit. We call this an open circuit, a circuit where there is a gap or a jump or no connection between two wires or parts of the circuit.

When the device is turned on by pressing the switch, it completes the circuit, the current starts flowing through it. We call this a closed circuit.

So, whenever a battery operated device is turned/switched on, it means the electrical circuit at that time is closed.

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How many liters of hydrogen gas are formed from the complete reaction of 1.03 mol of C? Assume that the hydrogen gas is collecte
aliina [53]

Answer:

27 liters of hydrogen gas will be formed

Explanation:

Step 1: Data given

Number of moles C = 1.03 moles

Pressure H2 = 1.0 atm

Temperature = 319 K

Step 2: The balanced equation

C +H20 → CO + H2

Step 3: Calculate moles H2

For 1 mol C we need 1 mol H2O to produce 1 mol CO an 1 mol H2

For 1.03 moles C we'll have 1.03 moles H2

Step 4: Calculate volume H2

p*V = n*R*T

⇒with p = the pressure of the H2 gas = 1.0 atm

⇒with V = the volume of H2 gas = TO BE DETERMINED

⇒with n = the number of moles H2 gas = 1.03 moles

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 319 K

V = (n*R*T)/p

V = (1.03 * 0.08206 *319) / 1

V = 27 L

27 liters of hydrogen gas will be formed

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What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2
atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

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