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3 years ago

Is oxygen a beginning substance or ending substance?

Chemistry

2 answers:

stiks02 [169]3 years ago

4 0

Answer:

Explanation:

The substances that are present at the beginning are called reactants and the substances present at the end are called products. Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products. Reactants are the starting materials, that is, whatever we have as our initial ingredients.Oxygen, a colorless, odorless, tasteless gas essential to living organisms, being taken up by animals, which convert it to carbon dioxide; plants, in turn, utilize carbon dioxide as a source of carbon and return the oxygen to the atmosphere. Oxygen forms compounds by reaction with practically any other element.

Vlad1618 [11]3 years ago

4 0

Answer:

<h2><u><em>A chemical reaction is the process in which atoms present in the starting substances rearrange to give new chemical combinations present in the substances formed by the reaction. These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products.</em></u></h2><h2><u><em></em></u></h2>

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Which statements correctly describe thermal energy in a block of ice, a glass of liquid water, and a container of steam?

DanielleElmas [232]

Answer:

Steam always has more thermal energy than ice.

Explanation:

6 0

4 years ago

Calculate δg∘rxn and e∘cell for a redox reaction with n = 3 that has an equilibrium constant of k = 4.4×10−2. you may want to re

lapo4ka [179]

a) First, to get ΔG°rxn we have to use this formula when:

ΔG° = - RT ㏑ K

when ΔG° is Gibbs free energy

and R is the constant = 8.314 J/mol K

and T is the temperature in Kelvin = 25 °C+ 273 = 298 K

and when K = 4.4 x 10^-2

so, by substitution:

ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)

= -7739 J = -7.7 KJ

b) then, to get E° cell for a redox reaction we have to use this formula:

ΔE° Cell = (RT / nF) ㏑K

when R is a constant = 8.314 J/molK

and T is the temperature in Kelvin = 25°C + 273 = 298 K

and n = no.of moles of e- from the balanced redox reaction= 3

and F is Faraday constant = 96485 C/mol

and K = 4.4 x 10^-2

so, by substitution:

∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)

= - 2.7 x 10^-2 V

8 0

3 years ago

explain the law of conservative of energy, give a specific example using kinetic and potential energy that shows how energy is c

KiRa [710]

Answer:

The Law of Conservation of Energy states that energy cannot be created or destroyed. In other words, the total energy of a system remains constant. This is an important concept to remember when dealing with energy problems. The two basic forms of energy that we will focus on are kinetic energy and potential energy.

Explanation:

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.

Im bad at these questions hope it helps and have a good day.

8 0

3 years ago

Serine has pka1 = 2.21 and pka2 = 9.15. use the henderson-hasselbalch equation to calculate the ratio neutral form/protonated fo

malfutka [58]

Calculate the ratio by using Henderson-Hasselbalch equation:

pH = pKa + log [neutral form] / Protonated form

3.05 = 2.21 + log [neutral form] / [Protonated form]

3.05 - 2.21 = log [neutral form] / [Protonated form]

0.84 = log [neutral form] / [Protonated form]

[neutral form] / [protonated form] = anti log 0.84 = 6.91

8 0

4 years ago

You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a

balandron [24]

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

6 0

3 years ago