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son4ous [18]
3 years ago
7

What does Dust in the atmosphere represents

Chemistry
1 answer:
Maksim231197 [3]3 years ago
7 0
Thats stars that have disapated
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How many copper atoms are in 150 g of copper metal
GaryK [48]
1. From grams -> mole:

=grams given x 1 mol/molar mass

So 150 g Cu x 1 mol Cu/63.546 g Cu = 2.4 mol Cu

2. From mole -> atoms

=number of mol x 6.022x10^22 atoms/1 mol

So 2.4 mol Cu x 6.022x10^22 atoms Cu/1 mol Cu = 1.4 x 10^24 atoms Cu
5 0
2 years ago
Molecules are the smallest units of matter with the characteristic properties of a substance.
Vilka [71]
False, Atoms are the smallest units of matter that display both the chemical and physical properties of it, based on the structure of the atom.
3 0
3 years ago
describe the behavior of the molecules in a liquid. Explain this behavior in terms of intermolecular forces.
tigry1 [53]

Answer:

Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions.

5 0
2 years ago
the decomposition of calcium carbonate to calcium oxide and carbon dioxide only takes place at very high temperatures, making th
USPshnik [31]
So calculate the H for the other two reactions a room temperature and combine the reactions to calculate the H of the decomposition of calcium carbonate using the Hess's Law
4 0
3 years ago
You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p
Elenna [48]

Answer:

  • <u>1.29 mol</u>

Explanation:

This is a direct application of the equation for ideal gases.

  • PV=nRT

Where:

  • P = pressure = 1.25 atm
  • V = volume = 25.2 liter
  • R = Universal constant of gases = 0.08206 atm-liter/K-mol
  • T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
  • n = number of moles

Solving for n:

  • n=\frac{PV}{RT}

Substituting:

n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol

8 0
3 years ago
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