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Ductility because ductility is the ability to be stretched and be made into wires.
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
Answer:it’s c
Explanation:just got it on edge
Explanation
Radius of neon atom : 69 pm = ![69\times 10^{-12} m](https://tex.z-dn.net/?f=69%5Ctimes%2010%5E%7B-12%7D%20m)
Volume occupied by the one atom:![\frac{4}{3}\pi r^3](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3)
![=\frac{4}{3}\times 3.14\times(69\times 10^{-12} m)^3=1.37\times 10^{-30} m^3](https://tex.z-dn.net/?f=%3D%5Cfrac%7B4%7D%7B3%7D%5Ctimes%203.14%5Ctimes%2869%5Ctimes%2010%5E%7B-12%7D%20m%29%5E3%3D1.37%5Ctimes%2010%5E%7B-30%7D%20m%5E3)
given that
atoms are present in 1L
1 L = 0.001 ![m^3](https://tex.z-dn.net/?f=m%5E3)
The volume occupied by the
neon atoms
![2.69\times 10^{22}\times 1.37\times 10^{-30} m^3=3.68\times 10^{-8} m^3](https://tex.z-dn.net/?f=2.69%5Ctimes%2010%5E%7B22%7D%5Ctimes%201.37%5Ctimes%2010%5E%7B-30%7D%20m%5E3%3D3.68%5Ctimes%2010%5E%7B-8%7D%20m%5E3)
Fraction of volume occupied by the neon atom:
![=\frac{3.68\times 10^{-8} m^3}{0.001 m^3}=3.68\times 10^{-11} m^3](https://tex.z-dn.net/?f=%3D%5Cfrac%7B3.68%5Ctimes%2010%5E%7B-8%7D%20m%5E3%7D%7B0.001%20m%5E3%7D%3D3.68%5Ctimes%2010%5E%7B-11%7D%20m%5E3)
![3.68\times 10^{-11} m^3](https://tex.z-dn.net/?f=3.68%5Ctimes%2010%5E%7B-11%7D%20m%5E3%3C0.001%20m%5E3%20%3D%201L)
The fraction of of volume occupied by the neon atom is very less than the 1 L which indicates the presence of large amount of empty space between the atoms of the gas.