Increase in the concentration of the reactants (vinegar and baking soda) leads to an increase in the rate of reaction (more volume of CO2 is evolved within a shorter time)

Explanation:

The chemical reaction between baking soda and vinegar in water is shown in the chemical reaction equation below;

NaHCO3(aq) + CH3COOH(aq) ----->CO2(g) + H2O(l) + CH3COONa(aq)

The chemical name of baking soda is sodium bicarbonate (NaHCO3) while vineager is a dilute acetic acid (CH3COOH) solution. This reaction provides a very easy set up in which we can study the effect of concentration on the rate of chemical reaction.

We must have it behind our minds that increase in the concentration of reactant species increases the rate of chemical reaction. Secondly, the rate of the reaction between baking soda and vinegar can be monitored by observing the volume of CO2 evolved and how quickly it evolves from the reaction mixture.

We can now postulate a hypothesis which states that; 'increase in the concentration of the reactants (vinegar and baking soda) leads to an increase in the rate of reaction (more volume of CO2 is evolved within a shorter time).'

If we go ahead to subject this hypothesis to experimental test, it will be confirmed to be true because a greater volume of CO2 will be evolved within a shorter time as the concentration of the reactants increases.

3 0

2 years ago

A solution that has a large amount of dissolved solute is called

kvasek [131]

Answer: It is an unsaturated solution

Explanation: This is because it has more solute than a normal solution.

6 0

2 years ago

1. Complete the following

Marat540 [252]

Answer:

$\large \boxed{\text{0.603 mol}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of O₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 18.1

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{18.1 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1005 mol C$_{6}$H$_{12}$O}_{6}$

b) Moles of O₂

$\text{Moles of O}_{2} =\text{0.1005 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol O}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.603 mol O}_{2}\\\\\text{The reaction requires $\large \boxed{\textbf{0.603 mol}}$ of oxygen}$

3 0

2 years ago