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3 years ago

Which of th following is true for a parallel circuit

Chemistry

1 answer:

emmainna [20.7K]3 years ago

8 0

Answer:

B. The voltage is the same across all resistors in the circuit.

Explanation:

The answer choices are:

A.The current is the same across all resistors in the circuit.

B.The voltage is the same across all resistors in the circuit.

C. As more resistors are added, the current will decrease.

D. The sum of the voltage drops will equal the total voltage.

<h2>Solution</h2>

In a simple parallel circuit, each component (resistor) is connected directly to the battery forming different branches or pathways to the current.

Thus, the ends of each resistor receives the total potential difference or voltage from the battery, which is what the statement B. asserts. Hence, statement B is true.

On the other hand, the current will split among the different pathways, following Ohm's law: the greater the resistance of the resistor the lower the current. Thus, the statement A is false.

For a parallel circuit, the total resistance decreases as more resistors are added, which means that, again following Ohm's law, the current will increase, contradicting the statement C. Thus, the statement C is false.

It is in the series circuit that 'the sum of the voltage drops will equal the total voltage"; in a parallel circuit the voltage drop is the just one and the same for each resistor. Thus, the statement D is false,too.

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What mass of propane could burn in 48.0 g of oxygen? C3H8 + 5O2 → 3CO2 + 4H2O

Ipatiy [6.2K]

Answer:

Mass = 13.23 g

Explanation:

Given data:

Mass of oxygen = 48.0 g

Mass of propane burn = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 48.0 g/ 32 g/mol

Number of moles = 1.5 mol

now we will compare the moles of propane and oxygen.

O₂ : C₃H₈

5 : 1

1.5 : 1/5×1.5 = 0.3 mol

Mass of propane burn:

Mass = number of moles × molar mass

Mass = 0.3 mol × 44.1 g/mol

Mass = 13.23 g

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4 years ago

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Read 2 more answers

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allochka39001 [22]

Answer:

$\large \boxed{\text{0.1767 mol/L}}$$

Explanation:

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3NaOH + H₃PO₄ ⟶ Na₃PO₄ + 3H₂O

(b) Moles of NaOH

$\text{Moles of NaOH} = \text{34.52 mL NaOH} \times \dfrac{\text{0.3840 mmol NaOH}}{\text{1 mL NaOH}} = \text{13.26 mmol NaOH}$

(c) Moles of H₃PO₄

The molar ratio is 1 mol H₃PO₄:3 mol NaOH.

$\text{Moles of H$_{3}$PO}_{4} = \text{13.26 mmol NaOH} \times\dfrac{\text{ 1 mmol H$_{3}$PO}_{4}}{\text{3 mmol NaOH}}\\\\= \text{4.419 mmol H$_{3}$PO}_{4}$

(d) Molar concentration of H₃PO₄

$c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V} = \dfrac{\text{4.419 mmol}}{\text{25.00 mL}} = \text{0.1767 mol $\cdot$ L$^{-1}$}\\\\\text{The molar concentration of the NaOH is $\large \boxed{\textbf{0.1767 mol/L}}$}$

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3 years ago