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Zinaida [17]
3 years ago
6

Which term is used when sugar is dissolved in water?a: covalentb: homogeneousc: amphibigeneousd: heterogeneous

Chemistry
1 answer:
Liula [17]3 years ago
4 0
Homogeneous because it completely dissolved so its now one. Heterogeneous would be when they don't dissolve and you see 2 separate substances like sand in water.
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What is the endpoint of a titration?Select one:When there is no acid and all base.When the amount of acid and base are equal.Whe
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Answer:

The correct option is: When the amount of acid and base are equal

Explanation:

Titration is an analytic method that is used to determine the concentration of an<em> unknown solution</em>, called <em>titrand</em>.

In this method, standard solution of known concentration, called <em>titrant</em>, is taken in the burette and added drop-wise to the titrand solution in the flask, until the endpoint is reached.

In case of an acid-base titration, a <em>pH indicator</em> is used, which changes the color of the solution when the endpoint is reached.

<u>The </u><u>endpoint</u><u> indicates the </u><u>equivalence point</u><u> of an acid-base titration, where the </u><em><u>concentration of the acid and base is equal</u></em><u>. </u>

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The powder could be acetaminophen, analgesic  having chemical formula C_{17}H_{21}NO_{4}

<h3>What is an empirical formula?</h3>

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.

We are given:

Percentage of C = 67.31 %

Percentage of H = 6.978 %

Percentage of N = 4.617 %

Percentage of O = 21.10 %

Let the mass of the compound be 100 g. So, the percentages given are taken as mass.

Mass of C = 67.31 g

Mass of H = 6.978 g

Mass of N = 4.617 g

Mass of O = 21.10 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of carbon = \frac{mass}{molar \;mass}

Moles of carbon =\frac{67.31g}{12g/mole}

=5.60 moles

Moles of hydrogen = \frac{mass}{molar \;mass}

Moles of hydrogen = \frac{6.978 g}{1 g/mole}

=6.978 moles

Moles of nitrogen =\frac{mass}{molar \;mass}

Moles of nitrogen = \frac{4.617  g}{14 g/mole}

=0.329 moles

Moles of oxygen =\frac{mass}{molar \;mass}

Moles of oxygen =\frac{21.10 g}{16 g/mole}

=1.31 moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.329 moles.

We get the ratio of C : H : N : O = 17 : 21 : 1 : 4

The empirical formula for the given compound is C_{17}H_{21}NO_{4}.

Learn more about the empirical formula here:

brainly.com/question/14044066

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