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beks73 [17]
3 years ago
8

Gold has a density of 19.3g/cm3. If you have a gold bar with a volume of

Chemistry
1 answer:
marin [14]3 years ago
8 0

Answer:

The answer is

<h2>877 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

density = 19.3 g/cm³

volume = 44.9 cm³

The mass of the gold bar

mass = 19.3 × 44.9 = 866.57

We have the final answer as

<h3>877 g</h3>

Hope this helps you

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A partir de 140 g de n2 y una cantidad de h2 se obtuvo 153G de NH³. Cual es el % de rendimiento de la reacción química
DiKsa [7]

Respuesta:

90.0 %

Explicación:

Paso 1: Escribir la ecuación química balanceada

N₂ + 3 H₂ ⇒ 2 NH₃

Paso 2: Calcular el rendimiento teórico de NH₃ a partir de 140 g de N₂

En la ecuación balanceada, participan de N₂: 1 mol × 28.01 g/mol = 28.01 g y de NH₃: 2 mol × 17.03 g/mol = 34.06 g.

140 g N₂ × 34.06 g NH₃ /28.01 g N₂ = 170 g NH₃

Paso 3: Calcular el rendimiento porcentual de NH₃

El rendimiento experimental de NH₃ es 153 g. Podemos calcular el rendimiento porcentual usando la siguiente fórmula.

R% = rendimiento experimental / rendimiento teórico × 100%

R% = 153 g / 170 g × 100% = 90.0 %

4 0
3 years ago
Indicate whether the following interactions of matter include a physical or chemical change. P/C? Evidence
Anarel [89]
1. CHEM CHANGE
2. PHYS CHANGE
3. CHEM CHANGE
4. CHEM CHANGE
5. PHYS CHANGE
3 0
3 years ago
2 Figure 3.1 shows two different forms of carbon, A and B.
raketka [301]

Answer:

show the pictures pls specify

5 0
3 years ago
PLEASE HELP WITH WORK INCLUDED
Mekhanik [1.2K]
I need more information?
7 0
3 years ago
What is the energy of light associated with a transition from n=3 to n=8 in a hydrogen atom? Does this represent absorption or e
solong [7]

Explanation:

Energy levels to be n = 8 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

1λ = R ⋅ (1/nf^2 − 1/ni^2)

where,

λ - the wavelength of the emitted photon

R - Rydberg's constant = 1.0974 x 10^7m

nf - the final energy = 8

ni - the initial energy level = 3

1/λ = 1.0974 x 10^7 * (1/8^2 − 1/3^2)

= -1.05x 10^6 m.

Using Heinsberg's equation,

E = (h * c)/λ

Calculating the energy of this transition you'll have to multiply Rydberg's equation by h * c

where,

h - Planck's constant = 6.626 x 10^−34 Js

c - the speed of light = 3.0 x 10^8 m/s

So, the transition energy, E = (6.626 x 10^−34 * 3 x 10^8) * -1.05x 10^6

= -2.08 x 10^-19 J.

B.

When an electron transitions from a less excited state to a excited state (higher energy orbit), the difference in energy is absorbed as a photon.

The energy is negative which means energy is lost or dissipated to the surroundings. Therefore, an absorption of photons.

8 0
3 years ago
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