<span>The
metallic properties that are caused by atoms rolling over each other in
metallic bonds are malleability and ductility.the atoms that are being rolled
over are delocalized electrons in the sea of electrons in the metallic bond
enable them to roll over when stress is applied.</span>
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Hello!
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❖ J.J Thomson was the first to discover atoms. He discovered atoms in 1897.
~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡
~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ
Given the model from the question,
- The products are: N₂, H₂O and H₂
- The reactants are: H₂ and NO
- The limiting reactant is H₂
- The balanced equation is: 3H₂ + 2NO —> N₂ + 2H₂O + H₂
<h3>Balanced equation </h3>
From the model given, we obtained the ffolowing
- Red => Oxygen
- Blue => Nitrogen
- White => Hydrogen
Thus, we can write the balanced equation as follow:
3H₂ + 2NO —> N₂ + 2H₂O + H₂
From the balanced equation above,
- Reactants: H₂ and NO
- Product: N₂, H₂O and H₂
<h3>How to determine the limiting reactant</h3>
3H₂ + 2NO —> N₂ + 2H₂O + H₂
From the balanced equation above,
3 moles of H₂ reacted with 2 moles of NO.
Therefore,
5 moles of H₂ will react with = (5 × 2) / 3 = 3.33 moles of NO
From the calculation made above, we can see that only 3.33 moles of NO out of 4 moles given are required to react completely with 5 moles of H₂.
Thus, H₂ is the limiting reactant
Learn more about stoichiometry:
brainly.com/question/14735801
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Hey there!
Molar mass NaCl = 58.44 g/mol
Number of moles
n = mass of solute / molar mass
n = 59.76 / 58.44
n = 1.0225 moles of NaCl
Volume in liters:
270 mL / 1000 => 0.27 L
Therefore:
M = number of moles / volume ( L )
M = 1.0225 / 0.27
= 3.78 M
Hope that helps!
Answer:
5.41 g
Explanation:
Considering:
Or,
Given :
For tetraphenyl phosphonium chloride :
Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)
Volume = 0.45 L
Thus, moles of tetraphenyl phosphonium chloride :
Moles of TPPCl = 0.01485 moles
Molar mass of TPPCl = 342.39 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass of TPPCl = 5.0845 g
Also,
TPPCl is 94.0 % pure.
It means that 94.0 g is present in 100 g of powder
5.0845 g is present in 5.41 g of the powder.
<u>Answer - 5.41 g</u>
