What is the balanced chemical symbol equation for silicon + water?

CO2+2H2O=H3O^++HCO3^-

bekas [8.4K]

Answer:

Bronsted Lowry Theory

$CO_2$ is the acid

$H_2 O$ is the base

$H_3 O^+$ is the conjugate acid and

$HCO_3^-$ is the conjugate base

Explanation:

Bronsted Lowry Theory:

An acid is a substance that can donate one or more protons

A base is a substance which can accept one or more protons

Hydrogen atom which is neutral (No Charge) contains 1 positive proton, 1 negative electron and 0 neutral neutron.

Thus Hydrogen atom has no Charge and it is neutral.

When an hydrogen atom loses an electron, Hydrogen ion is formed, which will contain 1 positive proton and 0 negative electron and 0 neutral neutron.

Thus Hydrogen ion has a positive charge.

Hydrogen ion is also called as a proton since it has only 1 proton in it.

Hydrogen ion in water that is,

$H^++H_2O\Rightarrow H_3O^+$

$H_3O^+$ is called as Hydronium ion.

Acid loses H+ (proton) to form conjugate Base

Base gains H+ (proton) to form conjugate Acid

For example

Let us consider the example given in the question

$CO_2+2H_2 O>H_3 O^+ +HCO_3^-$ can be written as (removed 1 $H_2 O$ from both the sides )

$CO_2+H_2 O>H^++HCO_3^-$ or

$CO_2+H_2 O > H_2 CO_3$ reversing the equation

$H_2 CO_3+H_2 O>H_3 O^+ +HCO_3^-$

$H_2 CO_3$ is the acid which donates $H^+$ to form $HCO_3^-$

$H_2 O$ is the base which gains $H^+$ to form $H_3 O^+$

$HCO_3^-$ is the conjugate base and $H_3 O^+$ is the conjugate acid

So

$CO_2$ is the acid

$H_2 O$ is the base

$H_3 O^+$ is the conjugate acid and

$HCO_3^-$ is the conjugate base

(Answer)

5 0

3 years ago

Describe two physical and two chemical changes involved in cooking. (4 points)

Ivenika [448]

Physical changes are when things get changed without altering chemical consistencies, which is melting solid butter into liquid one, or boiling water. Chemical changes are things such as caramelizing sugar when making sweets, or when carbon dioxide is created and released when baking bread.

7 0

4 years ago

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.

timofeeve [1]

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU + w

q = 475 J + 93 J

q = 568 J

Therefore, the magnitude of q for the process 568 J.

6 0

3 years ago

1.0 gallon to mm N.B 1 gallon=3.8L

I am Lyosha [343]

Answer:

Gallon (US) to Milliliter Conversion Table

Gallon (US) [gal (US)] Milliliter [mL]

1 gal (US) 3785.411784 mL

2 gal (US) 7570.823568 mL

3 gal (US) 11356.235352 mL

5 gal (US) 18927.05892 mL

4 0

2 years ago

This unbalanced equation represents a chemical reaction:

inn [45]

Answer:

2

Explanation:

Pb(NO3)2 + 2NaI 2NaNO3 + PbI2

5 0

3 years ago

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