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3 years ago

Why do atoms gain or lose electrons?

Chemistry

1 answer:

Aleksandr [31]3 years ago

6 0

Answer:kapil

Explanation:

Atoms lose electrons, if an electron gets more energy than then binding energy of the electron. This may be because of a collision with a charged particle or because of absorbtion of a photon. In a metal, there are just other positive charges nearby. The electron is not lost, but shared.

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The mass of a hypothetical planet is 1 100 that of the earth and it’s radius is 1 4 that of earth. If a person weighs 500 n on e

avanturin [10]

Answer: Your answer is 24

Explanation:

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2 years ago

PLS HELP!I will give brainliest! When producing hydrogen iodide, the energy of the reactants is 581 kJ/mol, and the energy of th

Veronika [31]

Answer:

Here, we are required to determine the total energy of the reaction and determine if the reaction is an endothermic or exothermic reaction.

The correct answer is option C.

First, we need to determine the energy of the reaction.

The energy of the reaction is the change in enthalpy between the product and reactants.

Change of Enthalpy,

Hreaction = Hproduct - Hreactant.

Therefore, for the reaction above, the change in enthalpy is:

Hreaction = 590kJ/mol - 581kJ/mol.

Hreaction = 9kJ/mol.

Hence, since the reaction has an enthalpy change of 9kJ/mol, the reaction is endothermic (i.e energy is absorbed).

Explanation:

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2 years ago

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

Write a variable expression that represents the following word phrase: three more than two times the difference of x and one.

Inga [223]

Convert the wordphrase to expression

$\\ \rm\longmapsto 2(x-1)+3$

$\\ \rm\longmapsto 2x-2+3$

$\\ \rm\longmapsto 2x+1$

5 0

3 years ago

Read 2 more answers

How many grams of O2 are needed to react with 18.2 g of NH3?

erastova [34]

Answer:

44 g oxygen are needed.

Explanation:

Given data:

Mass of oxygen needed = ?

Mass of ammonia = 18.2 g

Solution:

Chemical equation:

4NH₃ + 5O₂ → 4NO + 6H₂O

Now we will calculate the number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 18.2 g/ 17 g/mol

Number of moles = 1.1 mol

Now we will compare the moles of ammonia with oxygen from balance chemical equation.

NH₃ : O₂

4 : 5

1.1 : 5/4×1.1 = 1.375 mol

Mass of oxygen needed:

Mass = number of moles × molar mass

Mass = 1.375 mol × 32 g/mol

Mass = 44 g

4 0

3 years ago