Pascal’s Law states that when pressure is applied from an outside source to a contained fluid, the force is

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the de

Snezhnost [94]

Here is the full question

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.

Express your answer in two significant figures and include the appropriate units (in cm)

Answer:

ΔH ≅ 3.73 cm

Explanation:

The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:

P = ρ × g × h

where;

ρ is the density of the fluid

g is the gravitational constant

h is the height from the surface

From the question above;

For glycerine; we have:

density of glycerine = 1260 kg/m³

gravitational constant = 9.8 m/s²

height = ???

∴

$P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g$ ----- equation (1)

On the other hand for alcohol:

density of alcohol is given as = 790 kg/m³

gravitational constant = 9.8 m/s²

height = 10 cm

∴

$P_{(a)= 790kg/m^3*9.8m/s^2*10$ ----------- equation (2)

if we equate equation 1 and 2 together; we have

$P_{(g)= P_{(a)$

$1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm$

Making $h_g$ the subject of the formula, we have :

$h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}$

$h_g$ = 6.269 cm

The difference in the height denoted by ΔH can therefore be calculated as:

ΔH $= H_a-H_g$

ΔH $= 10cm - 6.269cm$

ΔH = 3.731 cm

ΔH ≅ 3.73 cm (to two significant figures)

5 0

4 years ago

H A and H B are both weak acids in water, and HA is a stronger acid than HB. Which of the following statements is correct? Selec

lubasha [3.4K]

Answer:

B is a stronger base than A^-, which is a stronger base than H2O, which is a stronger base than CI^-

Explanation:

The general equation for each acid is:

HA(aq) + H2O(ac) ⇄ H3O+(aq) + A-(aq)

HB(aq) + H2O(ac) ⇄ H3O+(aq) + B-(aq)

When these acids dissociate into its ions in water they lose a proton (H+), so they are proton donors (acids) and H2O is the proton acceptor (base). This reaction produces a conjugate acid and a conjugate base.

Conjugate base is what remains of the acid molecule after it loses a proton:

HA = acid A- Conjugate base

HB = acid B- Conjugate base

A conjugate acid is formed when the proton is transferred to the base

H2O = base H3O+ = Conjugate acid

The stronger acid will produce a weaker base. According to this, if HA is a stronger acid than HB, A- would be the weaker base (B- is the stronger base).

Compared with water, A- and B- are stronger bases because when they compete for a proton they have much greater affinity for H+ than water does and the equilibrium position will lie far to the left. (HA and HB are weak acids)

Finally Cl- is the weakest base because it comes after dissociation of HCl which is a strong acid

HCl(aq) + H2O → H3O+(aq) + Cl-(aq)

Note there is no double arrows, equilibrium lies far to the right. A strong acid yields a weak conjugate base it means one that has a low affinity for a proton.

4 0

3 years ago

How many molecules of water are in 1.52 moles of water?

abruzzese [7]

Answer:

9.15×10²³molecules

Explanation:

moles=number of particles/Avogadro's number

1.52=x/6.02×10²³

by cross multiplication;

x=1.52×6.02×10²³

=9.15×10²³

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5 0

3 years ago

If this same quantity of energy were transferred to 2.5 kg of water at it's boiling point what fraction of the water would be va

garik1379 [7]

Answer:

Fraction of water that can be vaporized = 0.0961 or 9.61%

Note : The question is incomplete. The complete question is given below:

A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body.

If this same quantity of energy were transferred to 2.5 kg of water at its boiling point, what fraction of the water would be vaporized?

Explanation:

Energy released by a serving of Cheeze-Its = 130 kcal

Since 1 kcal = 4.18 kJ, 130 kcal = 130 * 4.18 kJ = 543.4 kJ

Heat of vaporization (evaporating or condensing) Hv, of water = 2260 J/g

From formula, quantity of heat, Q = mass * heat of vaporization

mass of water = 2.5 kg = 2500 g

Q = 2500 g * 2260 J/g

Q = 5650000 J = 5650 kJ

Therefore 2.5 kg of water requires 5650 kJ of heat for complete vaporization

Fraction of water that can be vaporized by 543.4 kJ energy = 543.4/5650

Fraction of water that can be vaporized = 0.0961 or 9.61%

7 0

3 years ago

A 34.87 g sample of a substance is initially at 27.1 C. after absorbing 1071 J of heat, the temperature of the substance is 145.

Alenkinab [10]

$H= \frac{1071}{34.87}*145\\H= \frac{155295}{34.87} \\H= \frac{15529500}{3487} \\H=4453.542$

7 0

3 years ago