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3 years ago

Which of the following is a physical change?

Chemistry

1 answer:

Anit [1.1K]3 years ago

8 0

I’m pretty sure cooking an egg could be it

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Given the partial equation:

Nikolay [14]

Answer : The balanced chemical equation in acidic medium will be,

$IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$IO_3^-(aq)+Sn^{2+}(aq)\rightarrow I^-(aq)+Sn^{4+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-$

First balance the main element in the reaction.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-$

Now balance oxygen atom on both side.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-+6H^+\rightarrow I^-+3H_2O$

Now balance the charge.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}+2e^-$

Reduction : $IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O$

The charges are not balanced on both side of the reaction. Thus, we are multiplying oxidation reaction by 2 and the adding both equation, we get the balanced redox reaction.

Oxidation : $2Sn^{2+}\rightarrow 2Sn^{4+}+4e^-$

Reduction : $IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O$

The balanced chemical equation in acidic medium will be,

$IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)$

5 0

3 years ago

Design an invitation you could use to calculate the density of a penny

Nataliya [291]

Firstly, the density of any substance is represented by the mass (amount of matter) as divided by the volume(amount of space). According to external websites, the mass of a penny is 2.5 grams.However, the volume of a penny is .35cm to the power of 3 (due to the thickness of the penny being extremely minimal.Thus the amount of density is extremely little). Therefore, the density of a penny is 0.875 g/cm cubed (dimensional analysis).As for an invention that could be used, that is possible with the usage of a series of measurements that can both calculate mass and volume and directly allocate that to attain density

7 0

4 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

4 years ago

Chose the element pairings that are likely to react with each other Na and K or Na and Br

kaheart [24]

Answer:

Na k

Explanation:

because na is a metal and potassium is also a metal and both are active metal so is less likely to react as no bond is formed between them

7 0

3 years ago

An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equi

Arte-miy333 [17]

Answer:

713.51 N/m

Explanation:

Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.

From hook's law,

F = ke ...........................Equation 1

Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.

Make k the subject of the equation,

k = F/e ............................ Equation 2

Given: F = 264 N, e = 0.37 m.

Substitute into equation 2

k = 264/0.37

k = 713.51 N/m

Hence the spring constant of the bow = 713.51 N/m

3 0

3 years ago