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krek1111 [17]
3 years ago
9

certain conductor with a resistance of 10 Ohms is crossed by an electric current of intensity 100 mA. What is the potential diff

erence at the terminals of that conductor?
Chemistry
1 answer:
Morgarella [4.7K]3 years ago
8 0

Answer:

1.0 V

Explanation:

Step 1: Given data

  • Resistance of certain conductor (R): 10 Ω
  • Current's intensity (I): 100 mA
  • Potential difference at the terminals of that conductor (V): ?

Step 2: Convert "I" to Ampere

We will use the conversion factor 1 A = 1000 mA.

100 mA × 1 A/1000 mA = 0.100 A

Step 3: Calculate the potential difference

We will use Ohm's law.

V = I × R

V = 0.100 A × 10 Ω = 1.0 V

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Vikki [24]

Answer:

Cl-

Explanation:

Neutral (Cl) have 7 electron, but Cl- have 8 electron due to gain of 1 electron

4 0
1 year ago
I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even thou
Alexus [3.1K]

Answer:

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Explanation:

The reaction performed in the experiment is;

2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2

The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.

The oxidation - reduction equation is as follows;

2Cu2^+ + 2e ----> 2Cu^+ reduction half equation

2I^- ----> I2 + 2e. Oxidation half equation

Balanced redox reaction equation;

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

3 0
3 years ago
what is the limiting reactant when 1.50g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advan
PSYCHO15rus [73]
<h3>Answer:</h3>

Limiting reactant is Lithium

<h3>Explanation:</h3>

<u>We are given;</u>

  1. Mass of Lithium as 1.50 g
  2. Mass of nitrogen is 1.50 g

We are required to determine the rate limiting reagent.

  • First, we write the balanced equation for the reaction

6Li(s) + N₂(g) → 2Li₃N

From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.

  • Second, we determine moles of Lithium and nitrogen given.

Moles = Mass ÷ Molar mass

Moles of Lithium

Molar mass of Li = 6.941 g/mol

Moles of Li = 1.50 g ÷ 6.941 g/mol

                   = 0.216 moles

Moles of nitrogen gas

Molar mass of Nitrogen gas is 28.0 g/mol

Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol

                                     = 0.054 moles

  • According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
  • Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
  • On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.

Thus, Lithium is the limiting reagent while nitrogen is in excess.

7 0
3 years ago
HELP!!! Which element has 6 energy levels and 8 valence electrons?
Art [367]

Answer:

OXYGEN

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3 years ago
On molecular level what I causing the glaciers to melt
MakcuM [25]
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</span>That is because hydrogen bonds between water molecules in glaciers is decreased under increasing concentrations of carbon dioxide who <span>competes with the water molecules connected in the ice crystal.</span>
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3 years ago
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