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3 years ago

Other than temperature, which property also has an effect on the rates of chemical reactions?

1 answer:

8 0

Other than temperature, the property that also has an effect on the rates of chemical reactions is the concentration of the reactants. The correct answer is B

The rate of chemical reaction means how chemical react with respect to time or as a function of time. The rate of chemical reaction tells us the speed or how fast chemicals react.

There are many factors that can affect the rate of chemical reaction. Temperature is one of them. Increase in temperature will increase the rate at which chemical react.

The color of reactants, solubility of reactants and volume of reactants have nothing to do with the rate of chemical reaction.

Other than temperature, the property that also has an effect on the rates of chemical reactions is the concentration of the reactants.

Other factors affecting the rate of chemical reaction are:

the physical state of the reactants
surface area
temperature
and the presence of a catalyst

Therefore, the correct answer is option B

Learn more about chemical reaction here: brainly.com/question/11231920?source=archive

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Explain why it is dangerous to touch the live wire when a switch in the<br> mains circuit is open.

Fudgin [204]

Answer: all the energy is rushed in the wires due to the swich turning on when touching it may start a fire or electricute

Explanation:

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3 years ago

A 250 g air-track glider is attached to a spring with springconstant 4.0 N/m. Th damping constant due to air resistance is0.015

vaieri [72.5K]

Answer:

33.33 seconds

Explanation:

$N=\dfrac{1}{e}N_0$

$N_0$ = Initial length pulled = 20 cm

b = Damping constant = 0.015 kg/s

k = Spring constant = 4 N/m

m = Mass of glider = 250 g

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.25}{4}}\\\Rightarrow T=1.57079632679\ s$

Using exponential decay formula

$N=N_0e^{\dfrac{-bt}{m}}$

Final amplitude = Initial times decay

$\dfrac{1}{e}0.2=0.2e^{\dfrac{-0.015t}{2\times 0.25}}\\\Rightarrow 0.2=0.2e^{\frac{-0.015t}{2\cdot \:0.25}+1}\\\Rightarrow e^{\frac{-0.015t}{2\cdot \:0.25}+1}=1\\\Rightarrow \ln \left(e^{\frac{-0.015t}{2\cdot \:0.25}+1}\right)=\ln \left(1\right)\\\Rightarrow \left(\frac{-0.015t}{2\cdot \:0.25}+1\right)\ln \left(e\right)=\ln \left(1\right)\\\Rightarrow \frac{-0.015t}{2\cdot \:0.25}+1=\ln \left(1\right)\\\Rightarrow -\frac{0.015t}{0.5}=-1\\\Rightarrow -0.000225t=-0.0075\\\Rightarrow t=33.33\ s$

The time taken is 33.33 seconds

7 0

3 years ago

Estimate the kinetic energy of the earth with respect to the sun as the sum of two terms.

nekit [7.7K]

The definition of kinetic energy allows to find the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is $\frac{K_{Sum} }{K_{Earth}} = 5.3 \ 10^2$

<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

K = ½ m v²

where K is the kinetic energy, m the mass of the body and v the velocity of the body.

In a compound motion it is common to separate energy into parts to simplify calculations.

Translational kinetic energy. Due to the linear movement of the body

$K_{tras} =\frac{1}{2} m v^2$

Rotational kinetic energy. Due to the rotational movement of the body.

$K_{rot} = \frac{1}{2} I w^2$

Where I is the inrtia momentum and w the angular velocity.

They indicate that we compare the kinetic energy of the sun and the Earth.

The Earth has two movements, one of rotation about its axis with a period of T = 24 h and one of translation with respect to the Sun with a period of T= 365 days, therefore the kinetic energy of the Earth.

$K_{earth} = K_{tras} + K_{rot}$

Linear and rotational speed are related.

v = w r

The Earth is an almost spherical body therefore the moment of inertia of a solid sphere.

I = $\frac{2}{5 } m r^2$

Let's subatitute.

$K_{earth} = \frac{1}{2} \ m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}$

The movement of the Earth around the sun is almost circular, therefore we can use the relations of the uniform circular movement, where the angle for one revolution is 2π radians and the time is called the period.

$w = \frac{2 \pi}{T}$

Let's substitute.

$K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2 \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2$

$K_{earth} = 4 \pi^2 \ m \ ( \frac{1}{2} [ \frac{r_{tras}}{T_{tras}y} ]^2 + \frac{1}{5} [ \frac{r_{rot}}{T_{rot}}]^2)$

Data for Earth are tabulated:

Mass m = 5.98 1024 kg

Radius r = 6.37 10⁶ m

Radius orbits tras = 1.496 10¹¹ m

Rotation period $T_{rot}$ = 24 h ( $\frac{3600s}{1h}$ ) = 8.64 10⁴s

Translation period $T_{tras}$ = 365 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 3.15 10⁷ s

Let's calculate.

$K_{earth} = 4 \pi^2 5.98 \ 10^{24} ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2 \ + \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )$

$K_{earth} = 2.36 \ 10^{26 } \ (1.128 \ 10^7 + 1.087 \ 10^3)$

$K_{earth}= 2.66 \ 10^{33} J$

Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of $T_{Sum}$ = 27 days.