Answer:
Explanation:
To determine the molecular formula of the compound, the empirical formula must be determined first. To determine the empirical formula, the percentage of each constituent is divided by its molar mass. This is shown below
Carbon = 60/12 = 5
Oxygen = 32/16 = 2
Hydrogen = 8/1 = 8
The next step is to divide each ratio by the smallest value. The smallest value is 2. It becomes
Carbon = 5/2 = 2.5
It is approximated to 3
Oxygen = 2/2 = 1
Hydrogen = 8/2 = 4
Therefore, the empirical formula is
C3H4O
From the given relative molecular mass of the compound, the molecular formula can be determined
The only chemical that is a liquid at room temperature is Mercury. It's toxic, and has a high vapor pressure at room temperature.
Answer:
5 electrons
Explanation:
When an atom loses are gain the electrons ions are formed.
There are two types of ions.
Anion
Cation
1 = Anion
It is formed when an atom gain the electrons. when atom gain electron negative charge is created on atom. For example.
X + e⁻ → X⁻
2= Cation
It is formed when an atom loses the electrons. when atom lose electron positive charge is created on atom. For example.
X → X⁺ + e⁻
When an atom loses three electrons +3 charge is created. If this atom have 8 protons it means there were 8 electrons too. When it lost three electrons then remaining number of electrons are 5.
X → X³⁺ + 3e⁻
The correctanswer is D hope this helps
Answer:
Explanation:
From the information given:
no of moles of 
= 0.01 L × 0.0010 mol/L
no of moles of = 
no of moles of 
= 0.01 L × 0.00010 mol/L
no of moles of = 
Total volume = 0.02 L
![[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%20%5Cdfrac%7B1%5Ctimes10%5E%7B-5%7D%20%5C%20mol%7D%7B0.02%20%5C%20L%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%20%5C%5B%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%200.0005%20%5C%20mol%2FL)
![[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%281%5Ctimes%2010%5E%7B-6%7D%20%5C%20mol%29%7D%7B0.02%20%5C%20L%7D)
![[F^{-}] = 5 \times 10^{-5} \ mol/L](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%205%20%5Ctimes%2010%5E%7B-5%7D%20%20%5C%20mol%2FL)
![Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%200.0005%20%5Ctimes%20%285%5Ctimes%2010%5E%7B-5%7D%29%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%201.25%20%5Ctimes%2010%5E%7B-12%7D)
Since Q<ksp, then there will no be any precipitation of CaF2
