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Harrizon [31]
2 years ago
15

When completing the diffusion of a dye through water, where does the 0 mm mark on your ruler go?

Chemistry
1 answer:
Marina CMI [18]2 years ago
3 0

The Problem bother on the diffusion of substance

What is diffusion?

This can be defined as the movement of particles from a region of higher concentration to a region of lower concentration till equilibrium is reached

Explanation:

From the experiment seeks to measure the diffusion distance against time

Now when dye is dropped in water, the particles/molecules of the dye interacts with the molecules of water hence there is a mixtures which cause the dye to move from the initial position of higher concentration to lower concentration.

Hence the movement from the 0 mm mark increases in both directions

for more information about diffusion kindly visit

https://brainly.in/question/22196343

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Why is potassium iodine electronically neutral
AfilCa [17]
They would be neutral if their electonegativity difference was 0
8 0
3 years ago
Read 2 more answers
You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (d
Romashka-Z-Leto [24]

Answer:

66.67%

Explanation:

From the given information:

mass of cyclohexane = 2.9949 grams

density of cyclohexane = 0.779 g/mL

Recall that:

Density = mass/volume

∴

Volume = mass/density

So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL

= 3.8445 mL

Also,

mass of propylbenzene = 1.6575 grams

density of propylbenzene = 0.862 g/mL

Volume of propylbenzene =  1.6575 g/ 0.862 g/mL

= 1.9229 mL

The volume % composition of cyclohexane from the mixture is:

= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100

= (\dfrac{3.8445}{3.8445+1.9229})\times 100

= (\dfrac{3.8445}{5.7674})\times 100

= 66.67%

6 0
2 years ago
Complete the equation for the conversion of sucrose into glucose <br> (1)C12H22O11 + H2O
Svetlanka [38]

Answer:

C₁₂H₂₂O₁₁ + H₂O     →    C₅H₁₂O₆ + C₆H₁₂O₆

Explanation:

Chemical equation:

C₁₂H₂₂O₁₁ + H₂O     →    C₅H₁₂O₆ + C₆H₁₂O₆

Source of sucrose:

Sucrose is present in roots of plants and also in fruits. It is storage form of energy. Some insects and bacteria use sucrose as main food. Best example is honeybee which collect sucrose and convert it into honey.

Monomers of sucrose and hydrolysis:

Sucrose consist of monomers glucose and fructose which are join together through glycosidic bond. Hydrolysis break the sucrose molecule into glucose and fructose. In hydrolysis glycosidic bond is break which convert the sucrose into glucose and fructose. Hydrolysis is slow process but this reaction is catalyze by enzyme. The enzyme invertase catalyze this reaction.

The given reaction also completely follow the law of conservation of mass. There are equal number of atoms of elements on both side of chemical equation thus mass remain conserved.

8 0
3 years ago
What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
How many grams of water (H2O) are produced from 34 grams of oxygen (O2)? *
Dennis_Churaev [7]

Answer:

Explanation:

complete combustion reaction of  ethane is given by the reaction

2C2H6+7O2..............4CO2+6H2O

no of moles in 34 grams of O2=34/32=1.063

7mole of O2 produced 6 moles of H2O

therefore 1.063 moles of O2 produced=1.063*6/7=0.9 moles

now 0.9 moles of H2O contain how much grams=0.9*18=16.2 grams

3 0
2 years ago
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