Explanation:
Ozone, or 03,has two major resonance structures that contribute equally to the overall hybird structure of the molecule. The two structures are equivalent from the stability standpoint, each having a positive and a negative formal charge placed on two of the oxygen atoms.
Answer:
The molarity of the solution: 0,27M
Explanation:
First , we calculate the weight of 1 mol of NaCl:
Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol
58,5 g---1 mol NaCl
64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl
A solution molar--> moles of solute in 1 L of solution:
4 L-----1,09 mol NaCl
1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M
45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
The balanced equation for the reaction is given below:
2Al + 3Cl₂ —> 2AlCl₃
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
Therefore,
30 atoms of Al will require =
= 45 molecules of Cl₂.
Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
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D.
Misinterpretation of scientific knowledge for the influence of public opinion has been a repeating problem over the past, and usually creates a rift between the scientific community and the general public.
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is,
molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in
of ethane gas
And, as we know that
1 mole of ethane molecule contains
molecules of ethane
2.869 moles of ethane molecule contains
molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is,
molecule.