The _______ are found on the left side of the arrow in a chemical reaction. A. coefficients B. reactants C. products D. subscripts

Given data:

Density of iron (Fe) = 7.9 g/cm3

Length of one side of the iron cube = 1.64 * 10^2 cm

Now, the volume (V) of a cube in which the length of the side is 'a' cm is given as:

V = a^3

Volume of iron cube = (1.64 *10^2 cm)^3 = 4.41 * 10^6 cm3

The density (D) of an object of mass (m) and volume (V) is given as:

D = m/V

or, m = D*V

Therefore, mass of iron cube = 7.9 g/cm3 * 4.41 * 10^6 cm3

                                                = 34.84 *10^6 g

Answer:

1.25 g

Explanation:

Now we have to use the formula;

N/No = (1/2)^t/t1/2

N= mass of cesium-137 left after a time t (the unknown)

No= mass of cesium-137 present at the beginning = 5.0 g

t= time taken for 5.0 g of cesium-137 to decay =60 years

t1/2= half life of cesium-137= 30 years

Substituting values;

N/5= (1/2)^60/30

N/5= (1/2)^2

N/5= 1/4

4N= 5

N= 5/4

N= 1.25 g

Therefore, 1.25 g of cesium-137 will remain after 60 years.

Answer:

C) SO3

Explanation:

Lewis formula shows the bonding between atoms of a molecule and expresses the lone pair present in the atoms.

SO3 or Sulfur trioxide cannot be adequately described by a single Lewis formula because it has majorly 3 resonance structures because Sulfur does not follow the octet rule and can expand electrons in its outer shell.

Hence, the correct answer is C) SO3

Final answer:

SO3, or Sulfur Trioxide, is the molecule whose electronic structure requires the depiction of multiple resonance structures for adequate description. The rest of the species can be represented with a single Lewis structure.

Explanation:

The electronic structure that can only be described by drawing two or more resonance structures is SO3 (Sulfur Trioxide). This molecule has 24 valence electrons having a central atom with expanded octet. The Lewis structure is drawn in a way that three resonance structures are needed to represent the bonding in this molecule adequately. On the other hand, the rest of the species given in the options can be described using a single Lewis formula.

Learn more about Resonance Structures here:

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Answer:

Moles of hydrogen formed = 3.5 moles

Explanation:

Given that:-

Moles of aluminium= 4.0 mol

Moles of hydrogen bromide = 7.0 mol

According to the reaction:-

2 moles of aluminum react with 6 moles of hydrogen bromide

1 mole of aluminum react with 6/2 moles of hydrogen bromide

4 moles of aluminum react with (6/2)*4 moles of hydrogen bromide

Moles of hydrogen bromide = 12 moles

Available moles of hydrogen bromide = 7.0 moles

Limiting reagent is the one which is present in small amount. Thus, hydrogen bromide is limiting reagent. (7.0 < 12)

The formation of the product is governed by the limiting reagent. So,

6 moles of hydrogen bromide on reaction forms 3 moles of hydrogen

1 moles of hydrogen bromide on reaction forms 3/6 moles of hydrogen

7 moles of hydrogen bromide on reaction forms (3/6)*7 moles of hydrogen

Moles of hydrogen formed = 3.5 moles

Answer:

3.5 mol H2, HBr (limiting reactant)

Explanation:

4.0 mol Al × 3 mol H2/ 2 mol Al = 6.0 mol H2

7.0 mol HB ×3 mol H2/ 6mol HBr = 3.5 mol H2

Since 7.0mol of HBr will produce less H2 than 4.0mol of Al, HBr will be the limiting reactant, and the reaction will produce 3.5mol of H2.

Answer:

Stoichiometric coefficient of hydrogen gas is 1.

Stoichiometric coefficient of palmitic acid is 1.

Explanation:

Addition of hydrogen to double bond is termed as hydrogenation reaction.

According to stoichiometry, 1 mole of palmitoleic acid reacts with 1 mole of hydrogen gas to give 1 mole of palmitic acid.

Stoichiometric coefficient of hydrogen gas is 1.

Stoichiometric coefficient of palmitic acid is 1.

Answer:

a. endothermic reaction

Explanation:

In an endothermic reaction, heat is absorbed from the environment. This leaves the surrounding at a colder temperature compared to the system.

ΔH, the change in enthalpy is assigned a positive sign because the heat energy level of the final state is higher than that of the initial state.

Some examples are mostly dissolution substances in water.