Doesn't have to be a detailed explaination

Water sits in an open beaker. assuming constant temperature and pressure, the vapor pressure of the water ______________ as the

user100 [1]

For a given system, vapor pressure can be defined as the pressure exerted by the vapor phase in equilibrium with the condensed phase.

In this case, as water sits in an open beaker it tends to evaporate thereby forming water vapor. This vapor phase then exerts a pressure over the liquid water, which is termed as the vapor pressure of water at that temperature and pressure. As water continues to evaporate, more and more molecules escape into the vapor phase which thereby increases the vapor pressure.

Ans: The vapor pressure of water increases as the water evaporates.

3 0

3 years ago

The hormone adrenaline can affect only cells with

-Dominant- [34]

Answer:

appropriately shaped receptors

Explanation:

6 0

3 years ago

For+the+reaction+H2+++I2+-+2HI+the+equilibrium+constant,+kc+is+49+at+a+fixed+temperature.+Two+mole+of+hydrogen+and+two+moles+of+

Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M respectively.

Solution : Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially 0.11 0.11 C

At equilibrium (0.11-x) (0.11-x) (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0

3 years ago

Aluminum foil is often incorrectly termed tin foil. If the density of tin is 7.28g/cm³,what is the thickness of a piece of tin f

leonid [27]

Answer:

15.0 µm

Step-by-step explanation:

Density = mass/volume

D = m/V Multiply each side by V

DV = m Divide each side by D

V = m/D

Data:

m = 1.091 g

D = 7.28 g/cm³

l = 10.0 cm

w = 10.0 cm

Calculation:

(a) Volume of foil

V = 1.091 g × (1 cm³/7.28 g)

= 0.1499 cm³

(b) Thickness of foil

The foil is a rectangular solid.

V = lwh Divide each side by lw

h = V/(lw)

= 0.1499/(10 × 10)

= 1.50 × 10⁻³ cm Convert to millimetres

= 0.015 mm Convert to micrometres

= 15.0 µm

The foil is 15.0 µm thick.

4 0

3 years ago

i have a balloon that can hold 100 liters of air. if i blow up this balloon with 3 moles of oxygen gas at a pressure of 1.0 atmo

Tcecarenko [31]

Baloon with 3 moles og oxygen at 1 atm.The temperature of the balloon is 4 Kelvin.

An ideal gas is a theoretical gas composed of many randomly transferring factor particles that aren't difficult to interparticle interactions. the best gasoline idea is beneficial because it obeys the precise gas law, a simplified equation of country, and is amenable to evaluation under statistical mechanics.

An ideal gas is described as one for which both the extent of molecules and forces between the molecules are so small that they have got no effect at the behavior of the gas. The real gas that acts almost like a really perfect gasoline is helium. that is due to the fact helium, in contrast to maximum gases, exists as an unmarried atom, which makes the van der Waals dispersion forces as low as viable

Using the ideal gas equation:-

Given;

P₁ = 1 atm

V₁ = 100 L

n = 3

r = 8.314

T = PV/nR

= 1 × 100 / 3 × 8.314

= 4 K

Learn more about ideal gas here:-brainly.com/question/20348074

#SPJ4

4 0

1 year ago