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ivann1987 [24]
3 years ago
7

PLEASE ANSWER

Chemistry
1 answer:
yulyashka [42]3 years ago
4 0

Answer:

lower energy that helps the balance

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The number of water molecules present in a drop of water at room temperature​
g100num [7]
Water drops come in different sizes.
Let's imagine a drop weighs a quarter of a gram.
The molar mass of water is about 18g/mol, which means that 6.02 x 10^23 water molecules (AKA a mole of water molecules) weigh about 18 grams.
A quarter of a gram is 1/72 of 18, so it contains 1/72 times 6.02 x 10^23 molecules. That equals 8361111111111110000000 molecules.
In scientific notation that is... 8.36 x 10^21 molecules.
8 0
3 years ago
Draw and name the structures of the carboxylic acids and esters you put together using molecular models.
Neporo4naja [7]

Answer:

This question is incomplete

Explanation:

However, when the carboxylic acids and esters put together are obtained, the procedure below can be followed in drawing and naming them.

For carboxylic acid,

1) it should be noted that the functional group here is -COOH which is drawn as -C = OH

      |

     OH

2) The carbon of the functional group is included among the carbon to be counted when naming the structure. For example, the compound below is propanoic acid.

CH₃CH₂COOH - As you can see that there are 3 carbons linked chain there.

3) As can be seen in (2) above, the suffix "oic" is used to name carboxylic acids

4) The carbon chain here is saturated (meaning there is no double or triple bond within the carbon chain)

Example of a structure of carboxylic acid is

H₃C - CH₂ - CH₂ - C = OH

                              |

                             OH

The structure above is a butanoic acid

For ester

1)The functional group here is -COO- . which can be drawn as

- C = O

   |

  O -

(meaning one oxygen atom is double bonded to the carbon and the other oxygen atom is bonded to another carbon chain)

2) The alkyl group attached to the oxygen atom is first of all mentioned before the carbon chain attached from the left is mentioned. For example, CH₃CH₂CH₂COOCH₂CH₃ is ethyl butanoate

3) As seen from (2) above, the suffix "oate" is used to end the name of esters

4) As also seen from (2) above, the carbon of the functional group is also included while counting the carbon chain of the parent name (butanoate).

5) The carbon chains here are also saturated.

Example of this ester is

CH₃CH₂C = O

              |

             O - CH₂CH₃

The name of this compound is ethyl propanoate

3 0
3 years ago
Potassium hydroxide, also known as lye, dissociates into metal and hydroxide ions in water.
lesya692 [45]

Answer:

strong acid and weak base

Explanation:

#carryonlearning

5 0
3 years ago
g The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g) SO2(g) + Cl2(g) is first order in SO2Cl2 with a rate const
Alex_Xolod [135]

Answer : The concentration of SO₂Cl₂ after 720 min will be, 3.81\times 10^{-4}M

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.80\times 10^{-3}\text{ min}^{-1}

t = time passed by the sample  = 720 min

a = initial amount of the reactant  = 2.86\times 10^{-3}M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

720=\frac{2.303}{2.80\times 10^{-3}}\log\frac{2.86\times 10^{-3}}{a-x}

a-x=3.81\times 10^{-4}M

Therefore, the concentration of SO₂Cl₂ after 720 min will be, 3.81\times 10^{-4}M

5 0
3 years ago
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X , the freezing point of the solution is
lora16 [44]

Answer:

i=1.62 .

Explanation:

Let, i be the Van't Hoff Factor.

Moles of benzamide,=\dfrac{Mass}{molar\ mass}=\dfrac{70.4}{121.14}=0.58\ mol.

Molality of solution, m=\dfrac{moles  }{mass\ of\ solvent}=\dfrac{0.58}{0.85}=0.68\ molal.

Now, we know

Depression in freezing point, \Delta T=i\times K_f\times m  .....1

It is given that,

\Delta T=2.7^o C\\i=1 ( since\ it\ is\ non\ dissociable\ solutes)\\K_f  ( freezing\ constant)\\

Putting all these values we get,

K_f=3.949\ C/m.

Now, moles of ammonium chloride=\dfrac{70.4}{53.49}=1.316\ mol.

molality =\dfrac{1.316}{0.85}=1.54 molal.\\\Delta T=9.9 .

Putting all these values in eqn 1.

We get,

i=1.62 .

Hence, this is the required solution.

6 0
3 years ago
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