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alexira [117]
3 years ago
11

Metric equivalent to 4 tbsp

Chemistry
1 answer:
yulyashka [42]3 years ago
3 0
2 fl oz is the metric equivalent to 4 tbsp
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How did Ernest Rutherford change the atomic model?
nadya68 [22]

Answer:

D. He showed that most of an atom's mass was located in the atom's

nucleus.

Explanation:

Ernest Rutherford changed the atomic model because of his experiment which was the gold foil experiment. A beam of alpha particles was aimed at a piece of gold foil, most particles passed through but some were scattered backward which showed that the middle of an atom (nucleus) is the where most of the mass is located.

6 0
3 years ago
1.)Which is the correct order of the organization of the human body?
Darina [25.2K]
Answers:
1. A
2. D
3. B
6 0
3 years ago
Read 2 more answers
What is the main difference in covalent bonds in diatomic chlorine and diatomic oxygen? diatomic means two atoms of the same typ
Aneli [31]
The correct response is C. Chlorine forms 1 covalent bond while Oxygen forms 2 covalent bonds.
3 0
3 years ago
Un mol de amoniaco Tiene una masa molar de 17 g y ocupa un volumen de 22.4 l qué volumen ocupa el 50 g amoniaco en condiciones n
mr_godi [17]

Answer:

V  = 65.81 L

Explanation:

En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:

PV = nRT  (1)

Donde:

P: Presion (atm)

V: Volumen (L)

n: moles

R: constante de gases (0.082 L atm / mol K)

T: Temperatura (K)

De ahí, despejando el volumen tenemos:

V = nRT / P   (2)

Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:

n = 50 / 17 = 2.94 moles

Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:

V = 2.94 * 0.082 * 273 / 1

<h2>V = 65.81 L</h2>
4 0
3 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
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