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ArbitrLikvidat [17]
3 years ago
9

Which power source directs the electrons from oxidation-reduction reactions to flow through a device to give the device power? O

A. An electric motor B. A battery C. An electromagnet O D. An electric generator​
Chemistry
1 answer:
Lady bird [3.3K]3 years ago
4 0

Answer:

B. A battery.

Explanation:

I just took the test

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d

ionic bond is formed when there is transfer of electrons and is composed of a positive cation and a negative anion

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3 years ago
How many liters of oxygen are required to react completely with 14.8 mol of Al?
Bond [772]

Answer:

296 L  

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

                  4Al + 3O₂ ⟶ 2Al₂O₃

n/mol:        17.4

1. Moles of O₂

n = \text{17.4 mol Al}\times \dfrac{\text{3 mol O}_{2}}{\text{4 mol Al}}= \text{13.05 mol O}_{2}

2. Volume of O₂

You haven't given the conditions at which the volume is measured, so I assume it is at STP (0 °C and 1 bar).

At STP the molar volume of a gas is 22.71 L.

V = \text{13.05 mol}\times \dfrac{\text{22.71 L}}{\text{1 mol }}= \textbf{296 L}

8 0
3 years ago
Which of the following groups of materials would
Nikitich [7]
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4 0
3 years ago
Read 2 more answers
1. How many milliliters of 10.0 M HNO 3 are needed to prepare 0.350 L of 0.400 M solution?
tatyana61 [14]
<h3>Answer:</h3>

14 milliliters

<h3>Explanation:</h3>

We are given;

  • 10.0 M HNO₃

Prepared solution;

  • Volume of solution as 0.350 L
  • Molarity as 0.40 M

We are required to determine the initial volume of HNO₃

  • We are going to use the dilution formula;
  • The dilution formula is;

M₁V₁ = M₂V₂

Rearranging the formula;

V₁ = M₂V₂ ÷ M₁

    =(0.40 M × 0.350 L) ÷ 10.0 M

   = 0.014 L

But, 1 L = 1000 mL

Therefore,

Volume = 14 mL

Thus, the volume of 10.0 M HNO₃ is 14 mL

5 0
3 years ago
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