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Anna007 [38]
3 years ago
14

1. How many moles of potassium are present in 4.23 x 1025 potassium atoms?

Chemistry
1 answer:
eduard3 years ago
5 0

Answer:

70.26 moles

Explanation:

1 mole of potassium will contain 6.02 x 10²³ atoms of potassium

6.02 x 10²³ atoms of potassium  = 1 mole

4.23 x 10²⁵ atoms of potassium = 4.23 x 10²⁵ / 6.02 x 10²³

= 70.26 moles .

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A person's blood type is determined by the presence and absence of specific antigens on the cell membrane. Which of these would
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Answer:

C - no antibodies

Explanation:

I dont think there is any blood type without antibodies

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3 years ago
En que se fundamento Moseley para describir la tabla periódica actual?
Burka [1]

Answer:

La tabla periódica de moseley fue una manera de dar solución a esta problemática, ya que nos permite tener a la mano todos los elementos que se han descubierto hasta la actualidad y sus propiedades químicas más resaltantes.

A continuación se Describe una linea del tiempo de la creación de la tabla periodica:

1820 -  se crean las tríadas de Dobereiner.  

1863 - se crea el cilindro de chancourtois.  

1864 - se crean las octavas de newlands.  

1869 - se crea el primer prototipo de la tabla periódica.  

1914 - se fórmula la ley periódica moderna por Moseley.

Explanation:

4 0
3 years ago
Which of the characteristics below best describes organic compounds
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what are the options

4 0
3 years ago
Earth is approximately 1.5 x 10^8 km from the sun the. How far is earth from the sun meters
Zielflug [23.3K]

Answer:

{ \tt{1 \: km = 1000m}} \\ { \tt{1.5  \times {10}^{8} km =  (\frac{1.5 \times  {10}^{8} \times 1000 }{1} )m}} \\  = 1.5 \times  {10}^{11}  \: metres

4 0
2 years ago
A 20.0 g piece of aluminum at 5.00 C is dropped into 20.2 g of water at 90.00 C. The final temperature is 75.00 C. Use the First
bekas [8.4K]

Answer:

The specific heat of aluminium is 0.906 J/g°C

Explanation:

Step 1: data given

Mass of aluminium = 20.0 grams

Temperature = 5.00 °C

Mass of water = 20.2 grams

Temperature of water = 90.00 °C

The final temperature = 75.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: calculate the specific heat of aluminium

heat won = heat lost

Qaluminium = -Qwater

Q = m*c* ΔT

m(aluminium * c(aluminium) *ΔT(aluminium = -m(water) * c(water) *ΔT(water)

⇒with m(aluminium) = mass of aluminium = 20.0 grams

⇒with c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒with ΔT(aluminium) = the change of temperature = T2 - T1 = 75.00 °C - 5.00 °C = 70.00 °C

⇒with m(water) = the mass of water = 20.2 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = T2 - T1 = 75.00°C - 90.00 °C = -15.00 °C

20.0 * c(aluminium) * 70.00 = -20.2 * 4.184 * -15.00

c(aluminium) = 0.906 J/g°C

The specific heat of aluminium is 0.906 J/g°C

7 0
3 years ago
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