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ss7ja [257]
4 years ago
8

Need help in chemistry: how to convert kg to mg? For example: 0.00576kg converted to its equivalent measurement in mg? Can you s

how the process so I understand how to do it?
Chemistry
1 answer:
vesna_86 [32]4 years ago
5 0
1 kg ----------------- 1000000 mg
0.00576 kg -------- ?? 

0.00576 x 1000000 / 1 => 5760 mg

hope this helps!
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An atom is made of 5 protons, 6 neutrons, and 7 electrons. Another atom is made of 6 protons, 5 neutrons, and 3 electrons. What
koban [17]

Answer:

Explanation: Atoms are the building blocks of all matter. They consist of three sub-atomic particles: protons, neutrons and electrons. Protons and neutrons are found in the nucleus of an atom. Electrons are found in energy levels around the nucleus as shown in the diagram representing a carbon atom with 6 protons, 6 neutrons and 6 electrons.

Sub-atomic particles

Particle  Whereabouts in atom  Relative mass  Charge

Proton  Nucleus  1  +1

Neutron  Nucleus  1  0

Electron  Outside the nucleus  1/1840  -1

Diagram of carbon atom - nucleus with electrons orbiting

Diagram of carbon atom - nucleus with electrons orbiting

In chemistry we are particularly interested in electrons. As you will see later, this is because chemical reactions involve the rearrangement of electrons. Nuclei of atoms (protons and neutrons) usually remain unchanged (except in radioactive decay).

Electrons are arranged in atoms according to their energies. This is called the electronic structure or electronic configuration of the atom. A crude but still useful model says the electrons can be in different energy levels. Electrons in a particular energy level all have the same energy as one another. The lowest energy level can accommodate up to 2 electrons. The second level can accommodate up to 8 electrons. The third level can accommodate up to 18 electrons. The diagram shows the situation for a sulfur atom.

Electronic configuration of sulfur including energy level diagram

6 0
4 years ago
How many grams of oxygen are required to burn 13.5 g of acetylene
Vinil7 [7]

Answer:

41.54 grams of oxygen are required to burn 13.5 g of acetylene

Explanation:

The balanced reaction is:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • C₂H₂: 2 moles
  • O₂: 5 moles
  • CO₂: 4 moles
  • H₂O: 2 moles

Being the molar mass of the compounds:

  • C₂H₂: 26 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • C₂H₂: 2 moles* 26 g/mole= 52 grams
  • O₂: 5 moles* 32 g/mole= 160 grams
  • CO₂: 4 moles* 44 g/mole= 176 grams
  • H₂O: 2 moles* 18 g/mole= 36 grams

You can apply the following rule of three: if by stoichiometry 52 grams of acetylene react with 160 grams of oxygen, 13.5 grams of acetylene react with how much mass of oxygen?

mass of oxygen=\frac{13.5 grams of acetylene*160 grams of oxygen}{52 grams of acetylene}

mass of oxygen= 41.54 grams

<u><em>41.54 grams of oxygen are required to burn 13.5 g of acetylene</em></u>

<u><em></em></u>

8 0
3 years ago
In 2021, you were given a 100. g wine sample to verify its age. Using tritium dating you observe that the sample has 0.688 decay
nikdorinn [45]

Answer:

1984

Explanation:

Given the formula;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the radioactive isotope

t= age of the wine

Ao= initial activity of the wine

A= activity of the at time = t

0.693/12.3 = 2.303/t log (5.5/0.688)

0.693/12.3 = 2.079/t

0.056 = 2.079/t

t= 2.079/0.056

t= 37 years

The wine was produced 37 years ago which means that it was produced in the year 1984

7 0
3 years ago
Heyyyyyyyyyyyyyyyyyyyyyyyyyyy
mart [117]

Answer:

hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

Explanation:

7 0
3 years ago
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Which of the following is not an indication of a chemical reaction?
amm1812
Formation of a precipitate
4 0
4 years ago
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