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insens350 [35]
3 years ago
10

Which acid is naturally found in foods and can be safe to eat? sulfuric acid citric acid hydrochloric acid nitric acid

Physics
2 answers:
Hoochie [10]3 years ago
5 0

Answer:

Citric Acid.

Explanation:

Citric Acid is an Acid found in most citrus food.

For example:

Lemon, Orange and others.

Sulfuric Acid:

A strong acid, which means it's PH is or below 3.

Improper use of this acid may lead to death.

Eating and also touching this acid may lead to disaster.

This acid is not found in fruits, vegetables and also snacks from the food complex.

Can't be used as an ingredient for food making.

Causes Acidic rain.

We can see that it's really poisonous, it causes Acid to rain.

Nitric Acid:

A strong acid, which means it's PH is or below 3.

Improper use of this acid may lead to death.

Eating and also touching this acid may lead to disaster and even to death.

This acid is not found in fruits, vegetables and also snacks from the food complex.

Can't be used as an ingredient for food making.

Hydrochloric Acid:

A strong acid.

Can't be used as an ingredient for food making.

It's PH is or below 3. (Around 3)

Consuming it may lead to death.

It's found in our stomach. It helps to breakdown the food that we eat.

Even though it's a strong acid, it's found in our body, it can't harm us since it's found in moderate state, and is released when digesting food.

Hope this helps ;) ❤❤❤

Sergio [31]3 years ago
4 0

Answer:

\Huge \boxed{\mathrm{citric \ acid}}

Explanation:

Citric acid is naturally found in foods and can be safe to eat.

Citric acid is present in oranges, lemons, and other types of citrus fruits that we can eat.

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Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm  is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

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The difference is    U_f -U_i = 16 *10^{-7} J

Explanation:

From the question we are told that

     The radius of the soap bubble  is  R_o =  7.45 \ cm =  \frac{7.45}{100} =  0.0745 \ m

      The potential of the soap bubble is  V_1  =307.0 V

      The new radius of the soap bubble  is R_1 =  0.19 * 7.45=1.4155\ cm = 0.014155 \ m

The initial electric potential is mathematically represented as

     U_i  = \frac{V_1^2 R_o }{2k }

The final  electric potential is mathematically represented as

    U_f  = \frac{V_2^2 R_1 }{2k }

The initial potential is mathematically represented as

     V_1 =  \frac{kQ}{R_o}

The final  potential is mathematically represented as

        V_2 =  \frac{kQ}{R_1}

Now  

         \frac{V_2}{V_1}  =  \frac{R_o}{R_1}

substituting values

        \frac{V_2}{V_1}  =  \frac{7.45}{1.4155} =   \frac{1}{0.19}

=>      V_2 =  \frac{V_1}{0.19}

    So

         U_f  = \frac{V_1^2 R_2 }{2k * 0.19^2}

Therefore

        U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }

       U_f -U_i =     \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]

where k is the coulomb's constant with value 9*10^{9} \  kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

       U_f -U_i =     \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]

       U_f -U_i = 16 *10^{-7} J

           

     

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