What happens if the metal you throw in is MORE REACTIVE than the metal ion in solution?

blank is the energy required for a chemical reaction to take place, and blank is the energy associated with every substance.

Korolek [52]

Heat is the first one...and the secound one is kinetic

5 0

3 years ago

Read 2 more answers

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

vanillin is a flavoring agent in vanilla it has a mass percent composition of 61.15% c,5.3%H,31.55% o determine the empirical fo

Nutka1998 [239]

The empirical and molecular formulas will be $C_5H_5O_2$ and $C_1_0H_1_0O_4$ respectively.

<h3>Empirical and molecular formula</h3>

The compound contains C, H, and O.

C = 61.15/12 = 5.0958

H = 5.3/1 = 5.3

O = 31.55/16 = 1.9719

Divide by the smallest

C = 2.6

H = 2.7

O = 1

Thus, the empirical formula is $C_5H_5O_2$

Empirical formula mass = (12x5) + (1x5) + 16x2 = 97

n = 152.15/97 = 2

The molecular formula is $C_1_0H_1_0O_4$

More on molecular and empirical formulas can be found here: brainly.com/question/14425592

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6 0

2 years ago

A chemist dissolves of pure potassium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The t

Leya [2.2K]

Answer:

12.99

Explanation:

A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.

Step 1: Given data

Mass of KOH: 716. mg (0.716 g)

Volume of the solution: 130. mL (0.130 L)

Step 2: Calculate the moles corresponding to 0.716 g of KOH

The molar mass of KOH is 56.11 g/mol.

0.716 g × 1 mol/56.11 g = 0.0128 mol

Step 3: Calculate the molar concentration of KOH

[KOH] = 0.0128 mol/0.130 L = 0.0985 M

Step 4: Write the ionization reaction of KOH

KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)

The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M

Step 5: Calculate the pOH

We will use the following expression.

pOH = -log [OH⁻] = -log 0.0985 = 1.01

Step 6: Calculate the pH

We will use the following expression.

pH + pOH = 14

pH = 14 - pOH = 14 -1.01 = 12.99

8 0

2 years ago

. A standard dry cell has an output voltage of A. 1.5 VDC. B. 1.1 VDC. C. 1.2 VDC. D. 2.0 VDC.

NikAS [45]

The correct answer is letter A. 1.5 VDC. A standard dry cell has an output voltage of A. 1.5 VDC. Standard dry cell is a type of electricity-producing chemical cell, that is commonly use in households, and even portable devices. Dry cell is zinc-carbon cell and with nominal voltage of 1.5 volts.

4 0

3 years ago