The length of the arc of radius R and angle α is L = αR.
#pls help
tribal needle
pi radians = 180°
tardy pawn
Oh, interesting.
We have:
F(grav) = P = GmM/(R + h)^2
Thus:
P2/P1 = (R + h1)^2/(R + h2)^2 = ((R + h1)/(R + h2))^2
Since h1 = 0:
P2/P1 = (R/(R + h2))^2
Now, let's find h2.
R/(R + h2) = √(P2/P1)
(R + h2)/R = √(P1/P2)
1 + h2/R = √(P1/P2)
So, we get:
h = (√(P1/P2) - 1)R
Well, you need to substitute the ratio.
Note that your given ratio is P2/P1, but I used P1/P2 in the final expression.
Yes.
Yes, that's correct.
No.
Can you show your solution?
Well, can you type it?
Well, fine.
Let's find the mole fraction of C.
x(C) = n(C)/n(total) = n(C)/(n(C) + n(H))
n(C) = m(C)/M(C)
n(H) = m(H)/M(H)
So:
x(C) = (m(C)/M(C))/(m(C)/M(C) + m(H)/M(H)) = (150 g/(12 g/mol))/(150 g/(12 g/mol) + 50 g/(1 g/mol)) = 0.2 = 1/5
Thus:
x(H) = 1 - x(C) = 1 - 1/5 = 4/5
So, the ratio of carbon to hydrogen atoms in our compound is 1:4. So, it's methane.
Well, a line can only lie in zero, two or three quadrants.
For the three quadrants, the conditions are the following:
- Slope is neither 0 nor ∞.
- Line doesn't pass through the origin.
Yeah, that's a horizontal line.
No, A is a vertical line.
Well, have you simplified all the equations?
No, C passes through the origin.
Yes.
Can you write the simplified equations of each line?
Well, write all four.
Yes.
Hm... Don't know, honestly.
No, sorry, still not sure.
Oh, easy enough.
Solubilities (s) are given per 100 g of the solvent (H2O in our case, obviously). So, we make a proportion:
s(salt)/(100 g) = m(salt)/m(water)
As m(water) = m(sol.) - m(salt):
s(salt)/(100 g) = m(salt)/(m(sol.) - m(salt))
Now we find m(salt) from here.
(m(sol.) - m(salt))/m(salt) = 100 g/s(salt)
m(sol.)/m(salt) - 1 = 100g/s(salt)
m(sol.)/m(salt) = 100g/s(salt) + 1
m(salt) = m(sol.)/(100g/s(salt) + 1)
Now just substitute the values.
exotic sage
in which class are u?
tardy pawn
Hm, let's see. I will denote molality by b, though.
b(NaOH) = n(NaOH)/m(H2O) = c(NaOH)V(sol.)/(m(sol.) - m(NaOH)) = c(NaOH)V(sol.)/(ρ(sol.)V(sol.) - n(NaOH)M(NaOH))
Now, let's divide the numerator and denominator by V(sol.).
b(NaOH) = c(NaOH)/(ρ(sol.) - (n(NaOH)/V(sol.))M(NaOH)) = c(NaOH)/(ρ(sol.) - c(NaOH)M(NaOH))
And now we divide the numerator and denominator by c(NaOH).
b(NaOH) = 1/(ρ(sol.)/c(NaOH) - M(NaOH)) = 1/(1.12 kg/l/(3 mol/l) - 0.04 kg/mol) = 3 mol/kg
The answer I got is exact.
Maybe you rounded somewhere?
exotic sage
they are teaching u this in class 9th?
tardy pawn
they are teaching u this in class 9th?
Well, these are simple problems on solutions, so yeah, 9th grade seems alright for this.
exotic sage
Well, these are simple problems on solutions, so yeah, 9th grade seems alright f...
well no hes indian
tardy pawn
Oh, right... Well, then I don't know.
tribal needle
ar(FGDE) = 83 + 17 = 100
ar(EFD) = ar(DFG) = 100/2 = 50
ar(ADF) = ar(EFD) - ar(DEA)
ar(ADF) = 50 - 17 = 33
ar(ADF) = ar(ABF) + ar(DCF)
ar(ABCD) = ar(ADF) + ar(ABF) + ar(DCF)
ar(ABCD) = 33 × 2 = 66 cm²
exotic sage
ic
glacial flameBOT
@warm flume has given 1 rep to @tribal needle
tribal needle
urwelcome
tribal needle
the thing I observe is that
ar(EDB) = ar(EDC) = ar(AEB) = ar(AEC)
@mossy flicker how did u get b?
this
but you said it here?
mossy flicker
wow you guys are quick to respond