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3 years ago

Why is only one molecule of N2 needed to produce 2 molecules of NH3?

1 answer:

6 0

It's d, here's why.
Stoichiometry: N2 + H2 (for instance) --> NH3
There are 2 Nitrogen atoms (or parts) of Nitrogen on the left side of the equation, and 2 Hydrogen, and only one Nitrogen but three Hydrogen on the other side. Where did the extra Nitrogen go? Where did that Hydrogen come from? The answer is Stoichiometry.
N2 + H2 --> NH3 has to be balanced, so we add coefficients to the reactants and products, which indicate in what ratio they are consumed in the reaction. They effectively multiply the subscripts on the elements.
To balance Nitrogen, we have to add a 2 to the front of NH3, so we get 2NH3. Nitrogen is balanced, but Hydrogen isn't. There are now 6 Hydrogen being produced by the reaction, so we can add a 3 to the products side, making 3H2.
Now we have N2 + 3H2 --> 2NH3, and everything is balanced.

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The. bond dissociation enthalpies of the H-H bond and the H-Cl bond are 435 kJ mol^-1 and 431 kJ mol^-1, respectively. The ΔHfO

Novay_Z [31]

The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.

<h3>What is the dissociation enthalpy?</h3>

Given that;

H-H bond energy = 435 kJ mol^-1

H-Cl bond energy = 431 kJ mol^-1

ΔHfO of HCL(g) = -92kJ mol^-1

Bond dissociation enthalpy of the Cl-Cl bond = x

-92 = 435 + 431 + x

x = -92 - (435 + 431)

x = -958 kJ mol^-1

Learn More about dissociation enthalpy:brainly.com/question/9998007?

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6 0

2 years ago

What is the engine piston displacement in liters of an engine whose displacement is listed as 490 in^3?

marishachu [46]

Answer:

490 in^3 = 8.03 L

Explanation:

Given:

The engine displacement = 490 in^3

= 490 in³

To determine the engine piston displacement in liters L;

(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)

First, we will convert in³ to cm³

Since 1 in = 2.54 cm

∴ 1 in³ = 16.387 cm³

If 1 in³ = 16.387 cm³

Then 490 in³ = (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³

∴ 490 in³ = 8029.63 cm³

Now will convert cm³ to dm³

(NOTE: 1 L = 1 dm³)

1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm

∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³

If 1 cm³ = 1 × 10⁻³ dm³

Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³

≅ 8.03 dm³

∴ 8029.63 cm³ = 8.03 dm³

Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³

Since 1L = 1 dm³

∴ 8.03 dm³ = 8.03 L

Hence, 490 in³ = 8.03 L

3 0

3 years ago

Classify the following as either solutions or colloids. If a colloid, name the type of colloid and identify both the dispersed a

aev [14]

Answer:

a. glucose in water( solution)

b. smoke in air (colloids)

c. carbon dioxide in air (solution)

d. milk( colloids)

Explanation:

A solution is said to be formed when a solute dissolves in a solvent to form a homogeneous mixture. The solute particles are less than 10^-9m in size. Familiar solutions are those where the solute are dissolved in a liquid solvent. When the liquid water, the solution is known as an aqueous solution. A typical example is (glucose in water). In some other cases, the apparent solution of a solute in a solvent is accompanied by a chemical reaction and this is often known as a chemical reaction. A typical example is (carbon dioxide in air).

Colloids are also known as false solutions. Here, the individual solute particles are larger than the particles of the true solution, but not large enough to be seen by the naked eye. When a light beam is placed beside a beaker containing a colloid, the light rays of the beam can be clearly seen. This shows that it exhibits the Tyndall effect while a solution dosent exhibit such.

In a colloid, the liquid solvent is more appropriately know as the DISPERSION medium while the solid solute particles constitute the DISPERSED substance. This can either be solid, liquid or gas.

For example:

--> smoke in air : Dispersion medium is gas while the dispersed substance is solid.

--> milk: Dispersion medium is liquid while the dispersed substance is liquid.

7 0

3 years ago

Which of the following is a property of matter?

lana66690 [7]

Explanation:

Matter also exhibits physical properties. Physical properties are used to observe and describe matter. Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density.

5 0

3 years ago

Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p

Licemer1 [7]

N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
 moles is calculated by dividing mass by molar mass

mass of N2O4: 50.0 g
molar mass of N2O4 = 92.02 g/mol
molar mass of N2H4 = 32.05 g/mol.
mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of N2H4

 2)
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of N2H4 moles
N2H4 needed = 1,08 moles.
You have more that 1,08 moles N2H4, so this means the limiting reagent is not N2H4, it's N2O4. The molecule that has molecules that are left is never the limiting reactant.

3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
$moles= \frac{grams}{molar mass}$ (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams

4 0

3 years ago