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aivan3 [116]
3 years ago
15

Is it possible to prepare chlorine in lab without application of heat?​

Chemistry
1 answer:
marissa [1.9K]3 years ago
5 0

Answer:

i think it will explode

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Question 13 of 32
FromTheMoon [43]

Answer:

D

Explanation:

7 0
3 years ago
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A 25.00 mL sample of vinegar was titrated with 39.27 mL of 0.4293 M NaOH. Calculate the concentration of acetic acid in the vine
QveST [7]

Answer:

0.6743 M

Explanation:

HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O

First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:

  • Molarity = moles / volume
  • moles = Molarity * volume
  • 0.4293 M * 39.27 mL = 16.86 mmol NaOH

<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.

Finally we <u>calculate the concentration (molarity) of acetic acid</u>:

  • 16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M
4 0
3 years ago
An autoclave is used to sterilize surgical equipment because
avanturin [10]

The answer to your question is D!

5 0
3 years ago
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An aqueous solution of calcium hydroxide is standardized by titration with a 0.120 M solution of hydrobromic acid. If 16.5 mL of
Artist 52 [7]

<u>Answer:</u> The molarity of calcium hydroxide in the solution is 0.1 M

<u>Explanation:</u>

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HBr

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.120M\\V_1=27.5mL\\n_2=2\\M_2=?M\\V_2=16.5mL

Putting values in above equation, we get:

1\times 0.120\times 27.5=2\times M_2\times 16.5\\\\M_2=0.1M

Hence, the molarity of Ca(OH)_2 in the solution is 0.1 M.

7 0
3 years ago
1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of a
Archy [21]

Answer:

Explanation:

1)

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃  +   Al  →  3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

                               Al           :            Ag

                                1            :             3

                                0.223   :         3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

2)

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂  →   2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

                         HgO         :         Hg

                            2            :          2

                          3.12          :       3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

3)

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂  + 5O₂   →  2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

                 N₂O₅          :           O₂

                     2            :             5

                    12.99      :         5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

Mass = 1039.36 g

4)

Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆  + 15O₂   →  12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

                  C₆H₆         :           CO₂

                     2            :             12

                    0.103      :         12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g

3 0
3 years ago
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