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3 years ago

When two atoms of the same element have different numbers of neutrons and therefore different atomic masses, what are they?

1 answer:

8 0

All atomic elements consist of a positively charged nucleus surrounded by a cloud of negatively charged electrons. When two atoms of the same element have different numbers of neutrons and therefore a different atomic mass they are known as isotopes of the given element. An example of an isotope is deuterium, which is an isotope of hydrogen.

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Calculate the enthalpy of reaction for 2CO + O2 → 2CO2. given the following bond energies:

kotegsom [21]

Answer : The enthalpy change for the reaction is 1043 kJ/mol.

Explanation :

The given chemical reaction is:

$2CO+O_2\rightarrow 2CO_2$

As we know that:

The enthalpy change of reaction = E(bonds broken) - E(bonds formed)

$\Delta H=[(2\times B.E_{C\equiv O})+(1\times B.E_{O\equiv O})]-[2\times B.E_{C=O}]$

Given:

$B.E_{C\equiv O}$ = 1074 kJ/mol

$B.E_{O\equiv O}$ = 499 kJ/mol

$B.E_{C=O}$ = 802 kJ/mol

Now put all the given values in the above expression, we get:

$\Delta H=[(2\times 1074kJ/mol)+(1\times 499kJ/mol)]-[2\times 802kJ/mol]$

$\Delta H=1043kJ/mol$

Therefore, the enthalpy change for the reaction is 1043 kJ/mol.

7 0

3 years ago

A silver cube with an edge length of 2.38 cm and a gold cube with an edge length of 2.79 cm are both heated to 81.9 ∘C and place

never [62]

Answer:

The final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

Explanation:

Given data;

edge length of silver, a = 2.38 cm = 0.0238 m

edge length of gold, a = 2.79 cm = 0.0279 m

final temperature of silver, t = 81.9 ° C

final temperature of Gold, t = 81.9 ° C

initial temperature of water, t = 19.6 ° C

volume of water, v = 109.5 mL = 0.0001095 m³

Known data:

density gold 19300 kg/m³

density silver 10490 kg/m³

density water 1000 kg/m³

specific heat gold is 129 J/kgC

specific heat silver is 240 J/kgC

specific heat water is 4200 J/kgC

Calculated data

Apply Pythagoras theorem to determine the side of each cube;

Silver cube;

let L be the side of the silver cube

Taking the cross section of the cube (form a right angled triangle), the edge length forms the hypotenuse side.

L² + L² = 0.0238²

2L² = 0.0238²

L² = 0.0238² / 2

L² = 0.00028322

L = √0.00028322

L = 0.0168

Volume of cube = L³

Volume of the silver cube = (0.0168)³ = 4.742 x 10⁻⁶ m³

Gold cube;

let L be the side of the gold cube

L² + L² = 0.0279²

2L² = 0.0279²

L² = 0.0279² / 2

L² = 0.0003892

L = √0.0003892

L = 0.0197

Volume of cube = L³

Volume of the silver cube = (0.0197)³ = 7.645 x 10⁻⁶ m³

Mass of silver cube;

density = mass / volume

mass = density x volume

mass of silver cube = 10490 (kg/m³) x 4.742 x 10⁻⁶ (m³) = 0.0497 kg

Mass of Gold cube

mass of gold cube = 19300 (kg/m³) x 7.645 x 10⁻⁶ (m³) = 0.148 kg

Mass of water

mass of water = 1000 (kg/m³) x 0.0001095 (m³) = 0.1095 kg

Let the heat gained by cold water be Q₁

Let the heat lost by silver cube = Q₂

Let the heat lost by gold cube = Q₃

Let the final temperature of water = T

Q₁ = 0.1095 kg x 4200 J/kgC x (T–19.6)

Q₂ = 0.0497 kg x 240 J/kgC x (81.9–T)

Q₃ = 0.148 kg x 129 J/kgC x (81.9–T)

At thermal equilibrium;

Q₁ = Q₂ + Q₃

0.1095 x 4200 (T–19.6) = 0.0497 x 240 x (81.9–T) + 0.148 x 129 x (81.9–T)

459.9 (T–19.6) = 11.928 (81.9–T) + 19.092(81.9–T)

459.9T - 9014.04 = 976.9032 - 11.928T + 1563.6348 - 19.092T

459.9T + 11.928T + 19.092T = 9014.04 + 976.9032 + 1563.6348

490.92T = 11554.578

T = 11554.578 / 490.92

T = 23.54 ⁰C

Therefore, the final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

6 0

3 years ago

When 11.12 g of neon is combined in a 100 L container at 80oC with 9.59 g of argon, what is the mole fraction of argon?

PolarNik [594]

The correct answer for the question that is being presented above is this one: "0.3."

Here it is how to solve.
Molecular mass of Ar = 40
Molecular mass of Ne = 20
Number of moles of Ar = 9.59/40 = 0.239
Number of moles of Ne = 11.12/20= 0.556
Mole fraction of argon = 0.239/ ( 0.239 + 0.556) = 0.3

3 0

3 years ago

Which of the following is a true statement?

sergejj [24]

Answer:

Explanation:

Wind is geological therefore it is geological weathering

5 0

2 years ago

Why should you prefer dry ashing to wet ashing of food sample such as cabbage

julia-pushkina [17]

Answer:

Wet ashing is primarily used in the preparation of samples for subsequent analysis of specific minerals . It breaks down and removes the organic matrix surrounding the minerals so that they are left in an aqueous solution

Explanation:

8 0

2 years ago