Answer:
4.285 L of water must be added.
Explanation:
Hello there!
In this case, for this dilution-like problems, we need to figure out the final volume of the resulting solution so that we would be able to obtain the correct volume of diluent (water) to be added. In such a way, we can obtain the final volume, V2, as shown below:

Thus, by plugging in the initial molarity, initial volume and final molarity (0.587 M) we obtain:

It means we need to add:

Of diluent water.
Regards!
Answer:
The mass percentage of carbon can be found easily using the molar mass of C6H12O6, 180.1559 g/mol. We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass. C6H12O6 = CO2 + C3H6O3 + CH3OCH3 Take fructose for example. Compound.
Explanation: I looked it up
<span>Answer: a mixture.
</span><span>Justification:</span>
<span /><span>
</span><span>1) Pure substances have a definite chemical formula: the same kind of atoms with the same fixed ratios and chemical bonds. Therefore, the percents of each element do not varye.
</span>
<span /><span /><span>
2) Elements and compounds are pure substaces. For example, Fe, Mg, Ti, are elements, and CO₂, CO, H₂CO₃ are compounds. Each of them will have always the same kind of atoms, in the same ratio and with the same chemcial bonds. Therefore the percents of the elements do not varye.
</span><span />
<span>3) Mixtures are formed by the physical combination (not chemical bonds) of different elements or compounds in variable proportions. As indicated, this describes the material bronze, in virtue of the variation of its composition. Other examples of mixtures are solutions (like brines), air, ocean water, and milk: different brines, different oceans and different milk have different contents of elements or compounds.
</span><span>
</span>
Answer:
<h2><em>I hope this help you. Mark me as brainliest and rate please</em></h2>
Explanation:
<em>the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, Ka and pKa.
</em>
<em>
</em>
<em>It is important that you don't confuse the words strong and weak with the terms concentrated and dilute.
</em>
<em>
</em>
<em>As you will see below, the strength of an acid is related to the proportion of it which has reacted with water to produce ions. The concentration tells you about how much of the original acid is dissolved in the solution.
</em>
<em>
</em>
<em>It is perfectly possible to have a concentrated solution of a weak acid, or a dilute solution of a strong acid. </em>
An acid, example: “hydrochloric acid”