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love history [14]
3 years ago
9

Write a hypothesis about the effect of temperature and surface area on the rate of chemical reactions using this format: “If . .

. then . . . because. . . .” Be sure to answer the lesson question, “How do the factors of temperature and surface area affect the rate of chemical reactions?”
Physics
2 answers:
Paha777 [63]3 years ago
6 0

Answer:

If temperature and surface area increase, then the time it takes for sodium bicarbonate to completely dissolve will decrease, because increasing both factors increases the rate of a chemical reaction.

Explanation:

grandymaker [24]3 years ago
3 0
Effect of temperature.

"If the temperature of the substance is increased then the rate of chemical reaction is also increased because the kinetic energy is greater."

Effect of surface area.

"If the surface area is increased then the rate of reaction is increased because there will be more active sites for the reaction to occur. 
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A sample of an ideal gas initially occupies a volume of 6 L. The pressure of the sample is then doubled while it is cooled to on
Troyanec [42]

Answer:

V₂= 1 L

Explanation:

Given that

Volume occupies V₁= 6 L

Initial pressure = P₁

Initial temperature = T₁

The final pressure =P₂ = 2 P₁

Final volume =V₂

Final temperature = T₁/3

As we know that equation for ideal gas

P V = m R T

P=pressure,  V=volume,   T=temperature

m=mass  ,R=gas constant

Now from mass conservation

m=\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}

\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}

\dfrac{P_1\times 6}{RT_1}=3\times \dfrac{2P_1V_2}{RT_1}

6 = 3 x 2 V₂

V₂= 1 L

So the final volume will be 1 L

4 0
3 years ago
At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor
Irina18 [472]

Answer:

The charge flows in coulombs is

dq=1.843x10^{-5}C

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

I=N*\frac{dF}{Rdt}

\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}

dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}

N=1300, \beta=0.703, A=\pi*r^2=\pi*0.10^2=0.01\pi m^2, R=99.4+202=301.4Ω

\alpha_f=14.6,\alpha_i=165.4

Replacing :

dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}

dq=1.843x10^{-5}C

5 0
3 years ago
A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
vivado [14]

Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

54.66\ \text{revolutions}

Explanation:

\tau = Torque = 36.5 Nm

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

The wheel's moment of inertia is 21.6\ \text{kg m}^2

t = 60.6 s

\omega_i = 10.3 rad/s

\omega_f = 0

\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2

Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

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3 years ago
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