Answer:
P2=0.385atm
Explanation:
step one:
Given that the temperature T1= 60 Celcius
we can convert this to kelvin by adding 273k to 60 Celcius
we have T1= 333k
pressure P1= 0.470 atm
step two:
we know that the standard temperature is T2= 273K
Applying the temperature and pressure relationship we have
P1/T1=P2/T2
substituting our given data we have
0.47/333=P2/273
cross multiply we have
P2= (0.47*273)/333
P2= 128.31/333
P2=0.385 atm
Answer:
Q Line A
A: (a) 0 (b) 20
B: (b) 10 (c) 10
C: (c) 10 (d) 10
D: (d) 10 (e) 13
E: (e) 13 (f) 0
F: (f) 0 (g) 0
a: The velocity is whatever you need to see when you are specifically using a numberline so the answer for a is just 10
b: The velocity for this one is not as easy as the last one, its 7.
Explanation:
When you have a number line and your trying to find a numberline, you just have to subtract the smaller line to the largest number but if you are trying to find the velocity in beetween more than one thane you add the two biggest ones and subtract the smallest one and if you make a wrong move like add the biggest and smalles youll... still get the same answer so it doesnt matter really but its just easier to do the smallest one as the subtracting number just FYI. Happy spring break!
well, I'm not good in chemistry
but I think it's 5.5
Answer: Option (3) is the correct answer.
Explanation:
A molecule is polar when the net dipole moment of molecule is zero. When an electronegative atom is attached to an electropositive atom then the electronegative atom pulls the shared pair of electrons more towards itself.
As a result, the movement of electrons is more towards the electronegative atom and the dipole moment becomes zero. This will also induce partial positive and partial negative charge on electropositive and electronegative atom.
On the other hand, 
are covalent compounds and does not have a net zero dipole moment.
Thus, we can conclude that out of the given options 
represents a polar molecule.
Given the pH of the solution, we can determine [H⁺].
pH = -log[H⁺]
[H⁺] = 10⁻⁷·⁷⁸
[H⁺] = 1.659586907 x 10⁻⁸ M
We also know that pOH = 14 - pH. Thus, the pOH is 6.22. Given the pOH of the solution, we can determine [OH⁻].
pOH = -log[OH⁻]
[OH⁻] = 10⁻6.22
[OH⁻] = 6.02559586 x 10⁻⁷ M
The ionic product of water Kw, can then be calculated using the following equation.
Kw = [H⁺][OH⁻]
Kw = (1.659586907 x 10⁻⁸)(6.02559586 x 10⁻⁷)
Kw = 1.0 x 10⁻¹⁴
