All categories

3 years ago

How many atoms of San are in 0.796 moles of this element

Chemistry

1 answer:

xenn [34]3 years ago

8 0

Answer:

4.8× 10²³ atoms

Explanation:

Given data:

Number of moles of San element = 0.796 mol

Number of atoms present = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

0.796 mol × 6.022 × 10²³ atoms / 1mol

4.8× 10²³ atoms

You might be interested in

In the citric acid cycle, after the acetyl-group (2C) from acetyl-CoA is transferred to oxaloacetate (4C) to produce citrate (6C

Alborosie

Answer:

The carbons of the acetyl group oxidize which generate CO2, and in turn H2O.

Explanation:

The pyruvic acid that is generated during glycolysis enters the mitochondria. Inside this organelle, the acid molecules undergo a process called oxidative decaborxylation in which an enzyme of several cofactors is involved, one of which is coenzyme A. Pyruvic acid is transformed into an acetyl molecule and these are been introduced to the begining of the Krebs Cycle where the acetyl-group (2C) from acetyl-CoA is transferred to oxaloacetate (4C) to produce citrate (6C). As the molecule cycles the two carbons of the acetyl oxidize and are released in the form of CO2. Then the energy of the Krebs cycle becomes sufficient to reduce three NAD +, which means that three NADH molecules are formed. Although a small portion of energy is used to generate ATP, most of it is used to reduce not only the NAD + but also the FAD which, if oxidized, passes to its reduced state, FADH2

8 0

3 years ago

Many power plants produce energy by burning carbon-based fuels, which also produces carbon dioxide. Carbon dioxide is a greenhou

RUDIKE [14]

Answer:

a) 2.541 mol/MJ;

b) 1.124 mol/MJ;

c) 0.4354 mol/MJ;

d) 0.1835 mol/MJ

Explanation:

The enthalpy of formation (ΔH°f) is the enthalpy of a reaction to form a compound by its constituents. For CO₂, ΔH°f = - 393.5 kJ/mol.

The enthalpy of a reaction is the sum of the enthalpy of the products (each one multiplied by the number of moles) less the sum of the enthalpy of the reactants (each one multiplied by the number of moles). The ΔH°f for simple substances (with one atom) is 0. The combustion is the reaction between the fuel and the oxygen.

a) The combution reaction is:

C(s) + O₂(g) → CO₂(g)

ΔH°rxn = -393.5 kJ/mol = -393.5x10⁻³ MJ/mol

Number of moles per MJ released: 1/|ΔH°rxn|

n = 1/(393.5x10⁻³) = 2.541 mol/MJ

b) The combustion reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

H₂O is in the liquid state because it's at 1 atm and 25ºC.

ΔH°f, H₂O(l) = -285.3 kJ/mol

ΔH°f, O₂(g) = 0

ΔH°f, CH₄(g) = -74.8 kJ/mol

ΔH°rxn = [2*(-285.3 ) + 1*(-393.5)] - [1*(-74.8)]

ΔH°rxn = -889.3 kJ/mol = -889.3x10⁻³ MJ/mol

n = 1/889.3x10⁻³ = 1.124 mol/MJ

c) C₃H₈(g) + 10O₂(g) → 3CO₂(g) + 4H₂O(l)

ΔH°f,C₃H₈(g) = -25.2 kJ/mol

ΔH°rxn = [4*(-285.3) + 3*(-393.5)] - [1*(-25.2)]

ΔH°rxn = -2,296.5 kJ/mol = -2.2965 MJ/mol

n = 1/2.2965 = 0.4354 mol/MJ

d) C₈H₁₈(l) + (25/2)O₂(g) → 8CO₂(g) + 9H₂O(l)

ΔH°f, C₈H₁₈(l) = -250.1 kJ/mol

ΔH°rxn = [9*(-283.5) + 8*(-393.5)] - [1*(-250.1)]

ΔH°rxn = -5,449.4 kJ/mol = -5.4494 MJ/mol

n = 1/5.4494 = 0.1835 mol/MJ

4 0

3 years ago

Which chemical element represents the most abundant “atoms of an element” found in air?

zalisa [80]

Answer:

The order of elements by volume-fraction (which is approximately molecular mole-fraction) in the atmosphere is nitrogen (78.1%), oxygen (20.9%), argon (0.96%), followed by (in uncertain order) carbon and hydrogen because water vapor and carbon dioxide, which represent most of these two elements in the air, are variable.

I copied and pasted but I hope this information is helpful :)

3 0

3 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu

tatyana61 [14]

Answer:

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)

i mol 0.10 0.050 0.10

c mol -0.038 -0.038 +0019 +0.038

e mol 0.062 0.012 00.019 0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] = 0.062 M

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

5 0

3 years ago

What’s the answer to these 3 questions? thanks!

Citrus2011 [14]

Water has h bonding
H-H
Sodium fluoride
I think

4 0

3 years ago