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stiv31 [10]
2 years ago
10

The arrangement of electrons in the orbitals of an atom

Chemistry
1 answer:
MrRa [10]2 years ago
6 0
The answer is electron configuration. 
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The density of copper is 8.9 g/ml. How many Cm of copper are in 0.283kg of copper?
Natalka [10]
Solve from this pt on

6 0
2 years ago
A 14.3-cm3 sample of tin has a mass of 0.104 kg.
VladimirAG [237]
This is a true statement if it is density you are looking for... Density problem.....

Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits

The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.

Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.

The accepted value for the density of tin is 7.31 g/cm³
7 0
3 years ago
What is the atomic mass of<br> an atom?
Alborosie

Answer:

1 unit

Explanation:

It is expressed as a multiple of one-twelfth the mass of the carbon-12 atom, 1.992646547 × 10−23 gram, which is assigned an atomic mass of 12 units.

1/12 * 12 = 1. I think so.

4 0
2 years ago
HELP!!! WRITE THE CORRECT ISOTOPE NOTATION FOR EACH OF THE FOLLOW:
kakasveta [241]
The answer is 4 an atom
4 0
2 years ago
Se tiene una solución acuosa 2M de carbonato de potasio. Expresar su concentración en %p/v y Normalidad.
zepelin [54]

Answer:

Normalidad = 4N

%p/V = 27.6%

Explanation:

La solución 2M de carbonato de potasio contiene 2moles de carbonato por litro de solución. La normalidad son los equivalente de carbonato de potasio (2eq/mol) por litro de solución:

2moles * (2eq/mol) = 4eq / 1L = 4N

El porcentaje peso volumen es el peso de carbonato en gramos dividido en el volumen en mL por 100:

%p/V:

Masa K2CO3 -Masa molar: 138.205g/mol-

2moles * (138.205g/mol) = 276g K2CO3

Volumen:

1L * (1000mL/1L) = 1000mL

%p/V:

276g K2CO3 / 1000mL * 100

<h3>%p/V = 27.6%</h3>
6 0
2 years ago
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