Use a periodic table to help you answer these questions if needed.

Suppose a liquid level from 5.5 to 8.6 m is linearly converted to pneumatic pressure from 3 to 15 psi. What pressure will result

wlad13 [49]

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

5 0

4 years ago

What would be the answer

ser-zykov [4K]

Answer:

4

Explanation:

The balanced equation for the reaction is as follows:

4Al + 3O2 --> 2(Al2O3)

4 0

4 years ago

How many molecules of F2 react with 66.6 g NH3 ? (stoichiometry)

stiks02 [169]

<h3>Answer:</h3>

5.89 × 10^23 molecules of F₂

<h3>Explanation:</h3>

The equation for the reaction between fluorine (F₂) and ammonia (NH₃) is given by;

5F₂ + 2NH₃ → N₂F₄ + 6 HF

We are given 66.6 g NH₃

We are required to determine the number of fluorine molecules

<h3>Step 1: Moles of Ammonia </h3>

Moles = Mass ÷ Molar mass

Molar mass of ammonia = 17.031 g/mol

Moles of NH₃ = 66.6 g ÷ 17.031 g/mol

= 3.911 moles

<h3>Step 2: Moles of Fluorine </h3>

From the equation 5 moles of Fluorine reacts with 2 moles of ammonia

Therefore,

Moles of fluorine = Moles of Ammonia × 5/2

= 3.911 moles × 5/2

= 9.778 moles

<h3>Step 3: Number of molecules of fluorine </h3>

We know that 1 mole of a compound contains number of molecules equivalent to the Avogadro's number, 6.022 × 10^23 molecules

Therefore;

1 mole of F₂ = 6.022 × 10^23 molecules

Thus,

9.778 moles of F₂ = 9.778 moles × 6.022 × 10^23 molecules/mole

= 5.89 × 10^23 molecules

Therefore, the number of fluorine molecules needed is 5.89 × 10^23 molecules

5 0

3 years ago

A body of knowledge must be _____ in order to be considered science.

Law Incorporation [45]

The correct answer is realistic and reliable

5 0

3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago