Answer:
a) P = 9.58 psi for h=7.2 m
b) P=4.7 psi for h=5.94 m
Explanation:
Since the pressure Pon a static liquid level h is
P= p₀ + ρ*g*h
where p₀= initial pressure , ρ=density , g = gravity
then he variation of the liquid level Δh will produce a variation of pressure of
ΔP= ρ*g*Δh → ΔP/Δh = ρ*g = ( 15 psi - 3 psi) /( 8.6 m - 5.5 m) = 12/3.1 psi/m
if the liquid level is converted linearly
P = P₁ + ΔP/Δh*(h -h₁)
therefore choosing P₁ = 3 psi and h₁= 5.5 m , for h=7.2 m
P = 3 psi + 12/3.1 psi/m *(7.2 m -5.5 m) = 9.58 psi
then P = 9.58 psi for h=7.2 m
for P=4.7 psi
4.7 psi = 3 psi + 12/3.1 psi/m *(h -5.5 m)
h = (4.7 psi - 3 psi)/ (12/3.1 psi/m) + 5.5 m = 5.94 m
then P=4.7 psi for h=5.94 m
Answer:
C₁₀H₁₆O
Explanation:
Molecular formula can be determined from parent peak by using rule of thirteen.
Rule of Thirteen:
First divide the parent peak value by 13 as,
= 152 ÷ 13
= 11.69
Now, multiply 13 by 11,
= 13 × 11 (here, 11 specifies number of carbon atoms)
= 143
Now subtract 143 from 152,
= 152 - 143
= 9
Add 9 into 11,
= 9 + 11
= 20 (hydrogen atoms)
So, the rough formula we have is,
C₁₁H₂₀
Now, add one Oxygen atom to above formula and subtract one Carbon and 4 Hydrogen atoms as these numbers are equal to atomic mass of Oxygen atom as,
C₁₁H₂₀ -------O--------> C₁₀H₁₆O
Calculate Hydrogen deficiency index as,
HDI = (2x + 2 - y) / 2
where,
x = C atoms
y = H atoms
HDI = [2(10) + 2 - 16] / 2
HDI = (20 + 2 - 16) / 2
HDI = (22-16) / 2
HDI = 6 / 2
HDI = 3
It means, Camphor contains 1 double bond and two rings. As, HDI of double bond is 1 and HDI of one ring is 1.
Answer:
300 kPa
Explanation:
P2=P1V1/V2
You can check this by knowing that P and V at constant T have an inverse relationship. Hence, this is correct.
Answer:
197 grams
Explanation:
Without taking into account the number of sig figs you have for your values, you can say that the answer will once again be
197 g of aluminium bromide.
(I hope this is right)