The box will not move cuz both force vectors are equal in magnitude but exactly opposite in direction. So A is the correct option
Answer:
here you go
Explanation:
Atomic Number (Z) = Mass Number (A) - Number of Neutrons An atom has 5 protons, 5 electrons and 6 neutrons The atomic number = number of protons = number of electrons = 5 The mass number = 5 protons + 6 neutrons = 11
The answer is D solubility
hope this helps:)
Answer:
see explanations
Explanation:
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
Ci(NH₃) = 3.5mole/4L = 0.875M
Cf(NH₃) = 1.6mole/4L = 0.400M
Rate-1 => Δ[NH₃]/Δt = |(0.400M - 0.875M)/3min| = 0.158M/s
Rate-2 => 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt) => 6/4(0.158M/s) = 0.237M/s
Rate-3 => 5(Δ[NH₃]/Δt) = 4(Δ[O₂]/Δt) => 5/4(0.158M/s) = 0.237M/s
_________________________________________________________
NOTE: When setting up comparative rate expressions for a given reaction, set the rates expressions as equal then swap coefficient values. Then solve for rate of interest and substitute givens.
example: for NH₃ and H₂O
- set rates expressions equal => Δ[NH₃]/Δt = Δ[H₂O]/Δt
- then swap and insert coefficients from given rxn ...
- solve for rate of interest ...
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
=> 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt)
=> Δ[H₂O]/Δt = 6/4(Δ[NH₃]/Δt) = 6/4(0.237M/s) = 0.237M/s
Explanation:
A. 100°C to Kelvins
![T(K)=T(^oC)+273.15](https://tex.z-dn.net/?f=T%28K%29%3DT%28%5EoC%29%2B273.15)
![T(K)=100(^oC)+273.15=373.15 K](https://tex.z-dn.net/?f=T%28K%29%3D100%28%5EoC%29%2B273.15%3D373.15%20K)
B 600°R to Kelvins
![(T)^oK=((T)^oR)\times 1.8](https://tex.z-dn.net/?f=%28T%29%5EoK%3D%28%28T%29%5EoR%29%5Ctimes%201.8)
![(T)^oK=600\times 1.8 K = 1080 K](https://tex.z-dn.net/?f=%28T%29%5EoK%3D600%5Ctimes%201.8%20K%20%3D%201080%20K)
C. 98°F to Kelvins
![(T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}](https://tex.z-dn.net/?f=%28T%28K%29-273.15%29%3D%28T%28%5EoF%29-32%29%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D)
![(T(K))=(98(^oF)-32)\times \frac{5}{9}+273.15=309.81K](https://tex.z-dn.net/?f=%28T%28K%29%29%3D%2898%28%5EoF%29-32%29%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D%2B273.15%3D309.81K)
D. 77.4°F to degree Celsius
![((T)^oC)=((T)^oF-32)\times \frac{5}{9}](https://tex.z-dn.net/?f=%28%28T%29%5EoC%29%3D%28%28T%29%5EoF-32%29%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D)
![(T)^oC =(77.4^oF-32)\times \frac{5}{9}=25.22^oC](https://tex.z-dn.net/?f=%28T%29%5EoC%20%3D%2877.4%5EoF-32%29%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D%3D25.22%5EoC)
E. 77.4 K to degree Celsius
![T(^oC)=T(^K)-273.15](https://tex.z-dn.net/?f=T%28%5EoC%29%3DT%28%5EK%29-273.15)
![T(^oC)=77.4(K)-273.15=-195.75^oC](https://tex.z-dn.net/?f=T%28%5EoC%29%3D77.4%28K%29-273.15%3D-195.75%5EoC)
F. 77.4°R to degree Celsius
![(T)^oC=((T)^oR-491.67)\times \frac{5}{9}](https://tex.z-dn.net/?f=%28T%29%5EoC%3D%28%28T%29%5EoR-491.67%29%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D)
![(T)^oC=((77.4)^oR-491.67)\times \frac{5}{9}=-230.15 ^oC](https://tex.z-dn.net/?f=%28T%29%5EoC%3D%28%2877.4%29%5EoR-491.67%29%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D%3D-230.15%20%5EoC)