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Vinvika [58]
3 years ago
14

In chemical reactions, atoms are

Chemistry
1 answer:
Anna007 [38]3 years ago
5 0
<span>In chemical reaction, atoms always are rearranged. They are never destroyed or created because it would contradict the law of conservation of energy. The law of conservation of energy states that a matter cannot be created nor destroyed. It can only be formed in one way or the other. Also it cannot be neutralized. Resulting in one cannot be part of a chemical reaction. An example of this is the reaction HCl + NaOH -> NaCl + H2O. the reactants and products are not created, they are just formed into another substance.</span>
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When electrons flow steadily in one direction, the current is:
Sonbull [250]
The answer is direct
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3 years ago
2CrO4 2−(aq)+2H+(aq)⇌Cr2O7 2−(aq)+H2O(l)Write the equation for the equilibrium constant (K) of the reaction studied in this exer
Vikki [24]

Answer:

K = \frac {[Cr_{2}O_{7}^{2-}]}{[CrO_{4}^{2-}]^{2} \cdot [H_{3}O^{+}]^{2}}  

Explanation:  

The equilibrium constant for a given reversible aqueous reaction is defined by the product ratio of the concentrations between the products and reactants:      

                                           aA + bB ⇄ cC + dD

K = \frac {[products]^{p}}{[reactants]^{r}} = \frac {[C]^{c} \cdot [D]^{d}}{[A]^{a} \cdot [B]^{b}}  

<em>where K: is the equilibrium constant, [C] and [D]: are the product concentrations, [A] and [B]: are the reactant concentrations and a,b,c,d: are the stoichiometric coefficients from the reaction.   </em>

Therefore, based on the definition the equilibrium constant of our reaction is:    

2CrO₄²⁻(aq) + 2H₃O⁺(aq) ⇄ Cr₂O₇²⁻(aq) + 3H₂O(l)

K = \frac {[Cr_{2}O_{7}^{2-}]}{[CrO_{4}^{2-}]^{2} \cdot [H_{3}O^{+}]^{2}}  

Generally, the water concentration is omitted from the expressions.

I hope it helps you!

8 0
3 years ago
A student dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . the student notices that the vol
Troyanec [42]

Answer : The correct answer for molarity = 0.19 \frac{mol}{L} and Molality = 0.18 \frac{mol}{Kg}.

Given : Mass of aniline = 6.2 g

Volume of solvent = 350 mL Density of solvent = 1.04 g/mL

1) Molarity : It is defined as number of moles of solute present in Litre of solution . It is expressed as :

Molarity(M) = \frac{moles of solute(mole)}{volume of solution (L)}\

Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

Mole of solute = \frac{given mass of solute (g)}{molar mass of solute\frac{g}{1 mol}}

Molar mass of aniline (C₆H₅NH₂) = 93.13 \frac{g}{1 mol}

Mass of aniline = 6.2 g

Plugging values in mole formula :

Mole  =  \frac{6.2 g}{93.13 \frac{g}{1 mol}}

Mole = 0.0665 mol

Step 2 : To find volume of solution

Volume of solution = volume of solute + volume of solution

Since addition of aniline does not change final volume , so volume of solvent = volume of solution. Since volume is given in mL , so it need to be converted to L .

1 L = 1000mL

Volume of solution = \frac{350 mL}{1000mL}  * 1 L

Volume of solution = 0.350 L

Step 3: Plug value of mole and volume in molarity formula :

Molarity = \frac{0.0665 mol}{0.350 L }

Molarity = 0.19 M or 0.19 \frac{mol}{L}

------------------------------------------------------------------------------------------

2) Molality : It is defined as mole of solute present in Kilogram of solvent . It can be expressed as :

Molality (m) = \frac{mole of solute (mol)}{Kilogram of solvent (Kg)}

Following are the steps to calculate molality :

Step 1: To find mole of Solute :

Mole of solute can be found out using mole formula . It is same as done for molarity .

Mole = 0.0665 mol

Step 2 : To find kilogram of solvent :

Mass of solvent can be calculated using density formula as :

Density \frac{g}{mL} = \frac{mass (g) }{volume (mL)}

Plugging value in density formula :

1.04 \frac{g}{mL}  =  \frac{ mass }{350 mL}

Multiplying both side by 350 mL

1.04 \frac{g}{mL} * 350 mL = \frac{x}{350 mL } * 350 mL

Mass of solvent = 364 g

Since mass is in g, it need to be converted to Kg . ( 1 Kg = 1000 g )

Mass of solvent = \frac{364 g}{1000g} * 1 Kg

Mass of solvent = 0.364 Kg

Step 3: Plug values of mole and Kg in molality formula :

Molality = \frac{0.0665 mol}{0.364 Kg}

Molality = 0.18 m or 0.18 \frac{mol}{Kg}

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B

Explanation:

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A. glycerol
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