Why does it conduct more before and after this minimum point?

1 answer:

3 0

Let me give you an example. Lets say that we have an amount of Ba(OH)2 compared to H2SO4. And let's say that the Ba(OH)2 dissociates as Ba+2 + 2 OH-. H2SO4 dissociates as 2 H+ + SO4-2. So, what happens here is that w<span>hen the conductivity is at a minimum it means that stoichiometric amounts of Ba(OH)2 and H2SO4 are present and the only materials you can find there in the reaction vessel are H2O and BaSO4. That is why it conduct more before and after this minimum point</span>

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Chemical Reactions Unit Test

dusya [7]

The right side of the equation yields 3CO₂+6H₂O molecules if the 2 in front of 2O is changed to a 3.

The correct option is A.

<h3>What are molecules ?</h3>

The smallest identifiable unit into which a pure substance can be divided while still retaining its composition and chemical properties is called a molecule.

<h3>Briefing:</h3>

Given the chemical equation expressed as:

CH₄+2O₂ → CO₂ + 2H₂O

According to the equation, 1 mole of methane and 2 moles of oxygen combine to create 1 mole of carbon dioxide and 2 moles of water.

The right equation will be written as: If 3 moles of oxygen are used in place of the original amount of moles of oxygen.

3CH₄+2(3O₂) → 3CO₂ + 6H₂O

This shows that the right side of the equation will result in 3CO₂+6H₂O if the 2 in front of 2O is changed to a 3.

To know more about Molecules visit:

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The complete question is -

Use the equation to answer the question.

CH₄+2O₂ → CO₂ + 2H₂O

If you change the 2 in front of 2O₂ to a 3, what will be the change in the results on the right side of the equation?

A- There are now 3CO₂+6H₂O. molecules.

B- There is an extra O₂ molecule left over.

C- Nothing changes in the equation.

D- There are now 3H₂O molecules.

3 0

1 year ago

How many grams are there in 5 mol H20<br>Vour ans<br>BASE

xxTIMURxx [149]

Answer:

$\large \boxed{\text{90 g}}$

Explanation:

$\text{M$_{r}$ of H$_{2}$O} = 18.02\\\text{Mass} = \text{5 mol} \times \dfrac{\text{18.02 g}}{\text{1 mol}} = \text{90 g}\\\\\rm \text{There are $\large \boxed{\textbf{90 g}}$ in 5 mol of water.}$