Chemistry College

7. How many total atoms are present on the reactant side of the equation?
A. 3
B. 6
C. 7
D. 13

Answers

Answer 1

There are 6 atoms of oxygen present on the reactants side of the given reaction. Thus the correct answer is option B.

How do you calculate the total atoms present on the reactant side?

Firstly, let us write the accurate chemical equation and then subsequently we will balance it as and when needed.

4Fe+3O₂→2Fe₂O₃

This is the correct balanced chemical equation for the reaction. And as we can find that the amount of iron and oxygen atoms on the reactants side is equivalent to that of the products side. This corroborates that the given chemical equation is correctly balanced.

Now, the query is based on counting the total number of oxygen atoms present on the side of reactants. As we can clearly see on the reactant side  there are three Oxygen molecules that are present. That means that there are a total of six oxygen atoms, since 3*2 gives us a total of 6 oxygen atoms.

Hence, there are a sum total of six O₂ atoms that are present on the reactant side of the provided chemical equation.

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The full question is:

How many oxygen atoms are present on the reactant side of the chemical equation  4Fe+3O₂→2Fe₂O₃?

A. 3

B. 6

C. 7

D. 13

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Related Questions

A benzoic acid pellet weighing 6.54 g is placed in a bomb calorimeter along with 0.35 g fuse wire. The benzoic acid is ignited, and the temperature rise is 3.6°. What is the heat capacity of this calorimeter?

Answers

Answer:

C = -0.624 kJ/K

To calculate the heat capacity of the calorimeter, we need to use the following formula:

q = CΔT

where q is the heat released by the benzoic acid, C is the heat capacity of the calorimeter, and ΔT is the temperature rise observed.

First, we need to calculate the heat released by the benzoic acid:

q = m × ΔH

where m is the mass of the benzoic acid and ΔH is its heat of combustion.

The heat of combustion of benzoic acid is -3226 kJ/mol.

The molar mass of benzoic acid is 122.12 g/mol.

So, the number of moles of benzoic acid used in the experiment is:

n = m / M = 6.54 g / 122.12 g/mol = 0.0535 mol

The heat released by the combustion of the benzoic acid is:

q = n × ΔH = 0.0535 mol × (-3226 kJ/mol) = -172.6 kJ

The negative sign indicates that the reaction is exothermic (i.e., it releases heat).

Next, we need to calculate the heat absorbed by the calorimeter, which can be determined using the heat capacity equation:

q = C × ΔT

We know the value of q from the combustion of benzoic acid and the value of ΔT from the temperature rise observed, which is 3.6°C.

The mass of the fuse wire is not relevant to the calculation, so we can ignore it.

Substituting the values, we get:

-172.6 kJ = C × (3.6°C)

We need to convert the temperature rise to kelvin:

ΔT (K) = ΔT (°C) + 273.15 = 3.6°C + 273.15 = 276.75 K

Substituting this value, we get:

-172.6 kJ = C × (276.75 K)

Solving for C, we get:

C = -172.6 kJ / 276.75 K = -0.624 kJ/K

The heat capacity of the calorimeter is -0.624 kJ/K. The negative sign indicates that the calorimeter loses heat as the temperature rises, which is a common feature of bomb calorimeters.

Amount of CuO formed when 63.5 g of copper is heated strongly in air is:

Answers

The mass (in grams) of CuO formed when 63.5 g of copper, Cu is heated strongly in air is 79.5 g

How do I determine the mass of CuO formed?

First, we shall write the balanced equation for the reaction, This is given below:

2Cu + O₂ -> 2CuO

Now, we shall determine the mass of CuO formed from the reaction. Details below:

Molar mass of Cu = 63.55 g/molMass of Cu from the balanced equation = 2 × 63.55 = 127.1 g Molar mass of CuO = 79.55 g/molMass of CuO from the balanced equation = 2 × 79.55 = 159.1 g

From the balanced equation above,

127.1 g of Cu reacted to form 159.1 g of CuO

Therefore,

63.5 g of Cu will react to form = (63.5 × 159.1) / 127.1 = 79.5 g of CuO

Thus, the mass of CuO formed is 79.5 g

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Using the equations
2 Ha (g) + O2 (g) - > 2 H20 (1)
AH° = -572 kJ
Ca (s) + ½ O2 (g) -> CaO (s)
AH° = -635 kJ
CaO (s) + H,O (1) -> Ca(OH)2 (s) AH° = -64 kJ
Determine the enthalpy for the reaction
Ca (s) + 2 H,0 (1) -> Ca(OH), (S) + Ha (g).
-1271 kJ
-413 kJ
-985 kJ
-364 kJ
-285 kJ

Answers

The enthalpy for the following reaction Ca (s) + 2 H₂O → Ca(OH)₂(s) + H₂ (g) is -365kJ.

What is enthalpy?

A thermodynamic system's enthalpy, which is one of its properties, is calculated by adding the system's internal energy to the product of its pressure and volume. It is a state function that is frequently employed in measurements of chemical, biological, and physical systems at constant pressure, which the sizable surrounding environment conveniently provides. The pressure-volume concept describes the effort needed to create space for the system by displacing its surrounds in order to determine its physical dimensions.For solids and liquids under typical conditions, the pressure-volume term is relatively tiny, whereas it is only somewhat small for gases. As a result, in chemical systems, enthalpy serves as a stand-in for energy; for example, in chemistry, "energies" like bond, lattice, and solvation are actually differences in enthalpy.

The enthalpy for the reaction Ca (s) + 2 H₂O (1) -> Ca(OH)₂ (s) + H₂ (g) is -365 kJ. This can be calculated by using the calculated enthalpy values of the three given reactions above and the law of Hess's.

First, let's look at the given reactions:

2 H₂ (g) + O₂ (g) → 2 H₂0

ΔH° = -572 kJ

Ca (s) + ½ O₂ (g) →CaO (s)

ΔH° = -635 kJ

CaO (s) + H2O → Ca(OH)₂ (s)

ΔH° = -64 kJ

Now let's calculate the enthalpy for the reaction Ca (s) + 2 H2O (1) -> Ca(OH)₂ (s) + H₂ (g):

ΔH° = [2(-64 kJ) + (-572 kJ) + (-635 kJ)] - (-572 kJ + (-64 kJ + (-635 kJ))

ΔH° = -365 kJ

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Complete the w expression for the autoionization of water at 25 °C.

Answers

Answer:

Please mark brainlist

Explanation:

The equilibrium expression for this reaction is Kw = [H₃O⁺][OH⁻], where Kw is the autoionization constant for water. At 25°C, the value of Kw is 1.0 x 10⁻¹⁴.

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