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<span>Cl^- 1s^2 2s^2p^6 3s^2 3p^6 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 17- 10 =7 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6; 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 19- 10 = 9
</span>
S = 2 + 6.8 + 2.45 = 11.25
<span>Zeff(Cl^-) = 17 – 11.25 = 5.75 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6 same S as for Cl^- but Z increases by 2 hence </span>
<span>Zeff(K^+) = 19 - 11.25 = 7.75</span>
<span>Kinetic energy because it is taking the students to school.</span>
V₁(O2) = 6.50<span> L
</span>p₁(O2) = 155 atm
V₂(acetylene) = <span>4.50 L
</span>p₂(acetylene) =?
According to Boyle–Mariotte law (At constant temperature and unchanged amount of gas, the product of pressure and volume is constant) we can compare two gases that have ideal behavior and the law can be usefully expressed as:
V₁/p₁ = V₂/p₂
6.5/155 = 4.5/p₂
0.042 x p₂ = 4.5
p₂ = 107.3 atm
Answer:
mass of box 1 = 2.20 kg
mass of box 2 = 5.93 kg
Explanation:
Let the mass of box 1 and box 2 is respectively
and 
so we will have
Force applied on box 1 then acceleration
Now we know that contact force between them in above case is given as
now we have
Answer:
0.5m
Explanation:
v=f×lamda
v is 300m/s, f is 600Hz, lamda is ?
lamda=v/f
lamda=300/600
lamda =3/6=1/2m
