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Zepler [3.9K]
3 years ago
9

What is a product for 4 reaction type?

Chemistry
1 answer:
Crank3 years ago
6 0

Answer:

1= 2H₂ + O₂ → 2H₂O

2=CaCo₃ + heat → CaO +CO₂

3=CH₄ + 2O₂   → CO₂ +2H₂O

4=HCl + NaOH   → NaCl + H₂O

Explanation:

1 = Simple composition

The formation of water molecule is simple composition reaction. In this reaction two hydrogen atoms react with one oxygen atom and form one water molecules.

2H₂ + O₂ → 2H₂O

The amount of energy released is -285.83 KJ/mol. It is exothermic reaction.

2 = Simple decomposition reaction:

The break down of sodium hydrogen carbonate into sodium carbonate, carbondioxide and water is decomposition reaction. The decomposition reactions re mostly endothermic, because compound required energy to break.

2NaHCO₃ + heat → Na₂CO₃ + H₂O + CO₂

It is endothermic reaction.

Another example is:

CaCo₃ + heat → CaO +CO₂

3 = Combustion reaction

Consider the combustion of methane:

CH₄ + 2O₂   → CO₂ +2H₂O

The burning of methane is exothermic. The combustion reactions are exothermic because when fuel are burns they gives energy.

4 = Neutralization reaction

The neutralization reactions are those in which acid and base react to form the salt and the water. Some neutralization reactions are exothermic because they release heat. e.g

Consider the neutralization reaction of HCl and NaOH.

HCl + NaOH   → NaCl + H₂O

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Answer:

The atoms are separating, and sublimation is happening.

Explanation:

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Answer the lab question (“What is the effect of temperature on the solubility of a solid in a liquid?”) with a hypothesis:
valkas [14]

Explanation:

When experimenting, the best hypothesis to develop would be a null hypothesis (H₀). A null hypothesis is a statement indicating no change or effect.  In this case, it would be;

“There is no effect of temperature on the solubility of a solid in a liquid”

An alternative hypothesis (Hₐ) would be;

“There is an effect of temperature on the solubility of a solid in a liquid”

In this experiment, the null hypothesis would be rejected and the alternative would be accepted. This is because the experiment would show that increased temperatures of the liquid increases solubility of the solid in the liquid.

8 0
3 years ago
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The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc
Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles

b) moles of C_2H_4

\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 combine with 1 mole of H_2

Thus 0.33 mole of C_2H_4 will combine with =\frac{1}{1}\times 0.33=0.33 mole of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product.

As 1 mole of C_2H_4 give =  1 mole of C_2H_6

Thus 0.33 moles of C_2H_4 give =\frac{1}{1}\times 0.33=0.33moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g

Thus theoretical yield (g) of C_2H_6 produced by the reaction is 9.9 grams

7 0
3 years ago
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol
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Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻  (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

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3 years ago
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