<h2>Answer:</h2>
A) 3 atoms - 1 atom of Carbon and 2 atoms of oxygen.
B) 2 atoms of Nitrogen.
C) 6 atoms - 2 Carbon atoms, 2 Hydrogen atoms, and 2 Oxygen atoms.
<h2>Explanations:</h2>
A molecule is a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound. Molecules are made up of atoms.
According to the following information, we are to find the number of atoms in the given molecules.
A) For carbon dioxide CO₂, this molecule is made of 3 atoms - 1 atom of Carbon and 2 atoms of oxygen.
B) For the compound N₂, this molecule is made up of 2 atoms of Nitrogen.
C) For the compound CHCOOH, this molecule consists of 6 atoms - 2 Carbon atoms, 2 Hydrogen atoms, and 2 Oxygen atoms.
<span>1. Blocks Ultra violet rays that can be harmful.
2. Provides Oxygen, so we can live.
3. It blocks meteors from hitting Earth's surface.
</span><span>
</span>
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
<h3>How to calculate empirical formula?</h3>
The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.
The empirical formula of the given compound can be calculated as follows:
- Hafnium = 55.7% = 55.7g
- Chlorine = 44.3% = 44.3g
First, we convert mass values to moles by dividing by the molar mass of each element
- Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
- Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol
Next, we divide each mole value by the smallest
- Hafnium = 0.312 ÷ 0.312 = 1
- Chlorine = 1.25 ÷ 0.312 = 4
Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
Learn more about empirical formula at: brainly.com/question/14044066
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Atomic mass Sodium ( Na ) = 22.98 u.m.a
22.98 g ----------------- 6.02x10²³ atoms
175 g ------------------- ?? atoms
175 x ( 6.02x10²³) / 22.98 =
4.58x10²⁴ atoms of Na
hope this helps!
