All categories

3 years ago

Dissolving sodium chloride will

1 answer:

3 0

Water can dissolve salt because the positive part of water molecules attracts the negative chloride ions and the negative part of water molecules attracts the positive sodium ions. The amount of a substance that can dissolve in a liquid (at a particular temperature) is called the solubility of the substance.

You might be interested in

Calculate the change in entropy that occurs in the system when 3.10 mole of isopropyl alcohol (C3H8O) melts at its melting point

ozzi

Answer:

3.10 mole of C3H8O change in entropy is 89.54 J/K

Explanation:

Given data

mole = 3.10 moles

temperature = -89.5∘C = -89 + 273 = 183.5 K

ΔH∘fus = 5.37 kJ/mol = 5.3 ×10^3 J/mol

to find out

change in entropy

solution

we know change in entropy is ΔH∘fus / melting point

put these value so we get change in entropy that is

change in entropy 5.3 ×10^3 / 183.5

change in entropy is 28.88 J/mol-K

so we say 1 mole of C3H8O change in entropy is 28.88 J/mol-K

and for the 3.10 mole of C3H8O change in entropy is 3.10 ×28.88 J/K

3.10 mole of C3H8O change in entropy is 89.54 J/K

4 0

3 years ago

Read 2 more answers

Mercury is the 80th position in the periodic table how many protons does it have

snow_lady [41]

Errrrr, it has 80.

80 is the correct answer

4 0

3 years ago

Read 2 more answers

A computer base unit of mass 7.5 kg is dragged along a smooth desk. If the normal contact force is 23N and the tension in the ar

kramer

If there isn't any force then the normal contact force will be

N=m*g=7.5*9.81=73.58N

which is 73.58-23=50.58N less

so, there the person must pull at 23 degree upward

break down the tension in two components, vertical and horizontal.

vertical tension= 50.58=T*sin23

T=50.58/sin23=129.45N

7 0

3 years ago

Read 2 more answers

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

Real-world examples of power

Paha777 [63]

By definition, power is the amount of energy consumed (or produced) in a second. (or more precisely, it is the rate of change in energy).
so anything which uses energy in a known time period can be labeled with a power rating.

an example for power could be a nuclear plant; traditional nuclear plants produce somewhat close to 1 giga watts (which means 1 giga joules in a second)

3 0

3 years ago