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True [87]
3 years ago
14

Classify these substances as acidic, basic, or neutral: vinegar, baking soda, tomato juice, sugar. A) vinegar-acidic Baking soda

-basic Tomato juice-acidic Sugar-neutral B) vinegar-basic Baking soda-neutral Tomato juice-basic Sugar-acidic C) vinegar-neutral Baking soda-neutral Tomato juice-neutral Sugar-basic
Chemistry
1 answer:
Gnesinka [82]3 years ago
4 0

Answer:

A) vinegar-acidic Baking soda-basic Tomato juice-acidic Sugar-neutral.

Explanation:

Hello,

In this case, it is widely acknowledged that vinegar is a diluted and processed form of acetic acid which is of course acidic as well as tomato juice which has a pH of about 4.2. Baking soda, which is chemically known as sodium bicarbonate has a pH greater than 7, for that is reason it is basic. Finally, sugar, is known by its neutrality, for which its pH is about 7, for that reason it is neutral. In such a way, answer is A) vinegar-acidic Baking soda-basic Tomato juice-acidic Sugar-neutral.

Regards.

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vampirchik [111]

Answer: Oxygen is the limiting reagent.

Explanation: CS_2(l)+3O_2(g)\rightarrow CO_2(g)+2SO_2(g)

As can be seen from the given balanced equation:

3 moles of O_2 reacts with 1 mole of CS_2

1.52 moles of O_2 reacts with=\frac{1}{3}\times 1.52=0.51moles of CS_2

Thus O_2 is the limiting reagent as it limits the formation of products. (0.91-0.51)= 0.40 moles of CS_2 will remain as such and thus  CS_2 is an excess reagent.

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3 years ago
An erlenmeyer flask contains 15.00mL of 0.030 M HCI before titration. 5.00 mL of 0.050 of M NaOH is added to the HCI in the flas
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1.1

0.050*0.005

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8 0
3 years ago
A sample of an unknown compound has a percent composition of 52.14% carbon, 13.13% hydrogen, and 34.73% oxygen. Which compounds
Artemon [7]
<h3>Answer:</h3>

                The two possible compounds are;

                            1) Dimethyl Ether H₃C--O--CH₃

                           2) Ethanol  H₃C--CH₂--OH

<h3>Solution:</h3>

Step 1: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  =  52.14 ÷ 12.01

                      Moles of C  =  4.341 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  =  13.13 ÷ 1.01

                      Moles of H  =  13.00 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  =  34.73 ÷ 16.0

                      Moles of O  =  2.170 mol

Step 2: Find out mole ratio and simplify it;

                C                                        H                                     O

            4.341                                 13.00                              2.170

     4.341/2.170                      13.00/2.170                    2.170/2.170

               2                                      5.99                                    1

               2                                        6                                       1

Hence,  Empirical Formula  =  C₂H₆O

<h3>Result:</h3>

         As the molecular mass of compound is not given therefore, we can assume and guess the empirical formula to be the molecular formula. Hence, possible compounds are,

                            1) Dimethyl Ether H₃C--O--CH₃

                           2) Ethanol  H₃C--CH₂--OH

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