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3 years ago

You can increase the rate solute dissolves in solvent by______

2 answers:

5 0

Crushing particles of solute. The more surface area of the solute exposed to the solvent the faster the solute will dissolve and react with the solvent

geniusboy [140]3 years ago

5 0

Answer: Option (b) is the correct answer.

Explanation:

More is the number of solute particles present more will be the interaction take place between both the molecules of solute and solvent.

As a result, more will be the rate of reaction which means that when solute is added into the solvent after crushing it then it will readily dissolve into the solvent.

When we lower the temperature then there will occur decrease in kinetic energy of the particles. So, less number of collisions will take place between solute and solvent particles.

Hence, less will be the rate of reaction.

Thus, we can conclude that you can increase the rate solute dissolves in solvent by crushing particles of solute.

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One car has two and a half times the mass of a second car, but only half as much kinetic energy. When both cars increase their s

Leokris [45]

Answer:

$v_1 = 7.96 m/s$

$v_2 = 17.8 m/s$

Explanation:

Let the mass of the other car is "m" and its kinetic energy is

$K = \frac{1}{2}mv^2$

now the mass of the first car is two and half times and its kinetic energy is half that of other car

so we will have

$\frac{1}{2}(2.5m)v_1^2 = \frac{1}{2}(\frac{1}{2}mv^2)$

$2.5 v_1^2 = 0.5 v^2$

$v_1 = 0.447 v$

now speed of both cars is increased by value of 9 m/s

so now we will have same kinetic energy for both cars

$\frac{1}{2}(2.5 m)(0.447v + 9)^2 = \frac{1}{2}m(v + 9)^2$

$2.5(0.447 v + 9)^2 = (v + 9)^2$

$1.58(0.447v + 9) = v + 9$

$0.293v = 5.22$

$v = 17.8 m/s$

so speed of first car is

$v_1 = 0.447 v = 7.96 m/s$

$v_2 = 17.8 m/s$

3 0

3 years ago

Name 5 machine principles

lidiya [134]

The Wheel & Axle. The wheel has always been considered a major invention in the history of mankind

The Inclined Plane

The Wedge

The Pulley

The Screw

3 0

3 years ago

The crankshaft in a race car goes from rest to 3000 rpm in 3.0 s . (a) What is the crankshaft's angular acceleration?

Talja [164]

Answer:

75 rotations

Explanation:

f0 = 0, f = 3000 rpm = 50 rps, t = 3 s

(a) use first equation of motion for rotational motion

w = w0 + α t

2 x 3.14 x 50 = 0 + α x 3

α = 104.67 rad/s^2

(b) Let θ be the angular displacement

use second equation of motion for rotational motion

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 104.67 x 3 x 3

θ = 471.015 rad

The angle turn in one rotation is 2 π radian.

Number of rotation = 471.015 / (2 x 3.14) = 75 rotations

7 0

3 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is PV = C

T = C for isothermal process

Calculating for the work done in isothermal process

W = P₁V₁ $ln[\frac{P_{1} }{P_{2} }]$

= mRT₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵pV = mRT]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

ΔU = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = ΔU + W

= 0 + 270.588

= 270.588 kJ

4 0

3 years ago

Evaluate tan(249). O A. 1.36 B. 0.45 C. 0.41 D. 0.91

vodomira [7]

Answer:

tan 249 = 2.61

tan 249 = tan (249 - 180) = tan 69 = 2.61

3 0

2 years ago