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balu736 [363]
2 years ago
10

What is changes in a chemical reaction

Chemistry
1 answer:
marta [7]2 years ago
3 0
In a chemical change, the atoms in the reactants rearrange themselves and bond together differently to form one or more new products with different characteristics than the reactants. When a new substance is formed, the change is called a chemical change
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Draw the other possible resonance structure of each organic ion. In each case, draw the structure that minimizes formal charges.
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Three resonance structures  can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.

Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.

The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.

Learn more: brainly.com/question/4933048

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What is the molar mass of CO2
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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
2 years ago
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